Chapter 9 Indeterminate Forms-I

  • \(\dfrac{0}{0}\)
  • \(\infty - \infty\)
  • \(0^0\)
  • \(0 \times \infty\)
  • \(\dfrac{\infty}{\infty}\)
  • \(1^{\infty}\)
  • \({\infty}^0\)

are indeterminate forms.

9.1 Exercise 5

9.1.1 Question 1

Evaluate the limits:

  1. \(\lim_{x \to 2} \dfrac{x^3-2x^2 +2x-4}{x^2-5x+6}\)

This is of form \(\dfrac{0}{0}\), so using L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 2} \dfrac{3x^2 - 4x + 2}{2x -5}\\ &= \dfrac{12 -8 +2}{-1}\\ &= -6 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{\tan x - x}{x - \sin x}\)

This is also of form \(\dfrac{0}{0}\). So applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\sec^2 x -1}{1-\cos x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2\sec^2 x \tan x}{\sin x} \\ &= \lim_{x \to 0} \dfrac{2 \sec^2 x \sin x}{\sin x \cos x} = \lim_{x \to 0} 2 \sec^3 x\\ &= 2 \end{split} \end{equation*}\]

  1. \(\lim_{x \to a} \dfrac{x^n-a^n}{x-a}\)

This is also of form \(\dfrac{0}{0}\). So applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to a} \dfrac{nx^{n-1} - 0}{1}\\ &= na^{n-1} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{x - \sin^{-1} x}{\sin^3 x}\)

The expression can be written as, \[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{x - \sin^{-1} x}{x^3 \left(\dfrac{\sin x}{x}\right)^3}\\ \end{split} \end{equation*}\]

Using property of multiplication of two limits,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{x - \sin^{-1} x}{x^3 \times 1}, \text{ as } \lim_{x \to 0} \dfrac{\sin x}{x} = 1\\ &= \lim_{x \to 0} \dfrac{x - \sin^{-1} x}{x^3} \end{split} \end{equation*}\]

From the chapter on Taylor and Maclaurin series, we know the expansion of,

\[\begin{equation*} \begin{split} \sin^{-1} x &= x + \frac{x^3}{3!} + \frac{9x^5}{5!} + \ldots \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{x - \sin^{-1} x}{x^3}\\ &= \lim_{x \to 0} \dfrac{x - \left(x + \frac{x^3}{3!} + \frac{9x^5}{5!} + \ldots\right)}{x^3}\\ &= \lim_{x \to 0} \dfrac{- \frac{x^3}{3!} - \frac{9x^5}{5!} - \ldots}{x^3} \Longrightarrow \left(\dfrac{0}{0}\right)\\ \end{split} \end{equation*}\]

Applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{-\dfrac{3x^2}{6} - \dfrac{45x^4}{120}}{3x^2}\\ &= \lim_{x \to 0} \dfrac{-\dfrac{3}{6} - \dfrac{45x^2}{120}}{3}\\ &= \dfrac{-\dfrac{3}{6}}{3}\\ &= -\dfrac{1}{6} \end{split} \end{equation*}\]

Alternative method

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{x - \sin^{-1} x}{x^3} \Longrightarrow \left(\dfrac{0}{0}\right)\\ \end{split} \end{equation*}\]

Applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{1 -(1-x^2)^{-1/2}}{3x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{1}{2}(1-x^2)^{-3/2}\times (-2x)}{6x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-(1-x^2)^{-3/2}}{6}\\ &= -\dfrac{1}{6} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{2 \sin x - \sin 2x}{\tan^3 x}\)

This can be written as,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{2 \sin x - \sin 2x}{x^3 \times \left(\dfrac{\tan x}{x}\right)^3}\\ &= \lim_{x \to 0} \dfrac{2 \sin x - \sin 2x}{x^3 \times 1}\\ &= \lim_{x \to 0} \dfrac{2 \sin x - \sin 2x}{x^3} \Longrightarrow \left(\dfrac{0}{0}\right)\\ \end{split} \end{equation*}\]

Applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{2\cos x - 2\cos 2x}{3x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-2 \sin x + 4 \sin 2x}{6x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-2 \cos x + 8\cos 2x}{6}\\ &= \dfrac{6}{6} = 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 1} \dfrac{1 + \log x -x}{1 - 2x + x^2}\)

This is also of form \(\dfrac{0}{0}\). So applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 1} \dfrac{\dfrac{1}{x} - 1}{-2 + 2x} \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 1} \dfrac{-x^{-2}}{2}\\ &= - \dfrac{1}{2} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{\log (1-x^2)}{\log \cos x}\)

This is also of form \(\dfrac{0}{0}\). So applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\dfrac{1}{1-x^2}\times (-2x)}{-\tan x} \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{2(1-x^2) + 4x^2}{(1-x^2)^2}}{\sec^2 x}\\ &= 2 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{xe^x - \log (1+x)}{x^2}\)

This is also of form \(\dfrac{0}{0}\). So applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{xe^x + e^x - \dfrac{1}{1+x}}{2x}\\ \end{split} \end{equation*}\]

Again of form \(\dfrac{0}{0}\), so

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{xe^x + e^x + e^x + \dfrac{1}{x^2}}{2}\\ &= \dfrac{0 + 1+ 1+ 1}{2}\\ &= \dfrac{3}{2} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{\tan x -x}{x^2 \tan x}\)

The expression can be written as,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\tan x -x}{x^3\dfrac{\tan x}{x}}\\ &= \lim_{x \to 0} \dfrac{\tan x -x}{x^3}, \text{ as } \lim_{x \to 0} \dfrac{\tan x}{x} = 0\\ &= \lim_{x \to 0} \dfrac{\tan x -x}{x^3} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \dfrac{\sec^2 x -1}{3x^2} & \Longrightarrow \left(\dfrac{0}{0}\right) \\ &= \dfrac{2\sec^2 x \tan x}{6x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \dfrac{\sec^2 x \tan x}{3x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \dfrac{\sec^4 x + 2 \sec^2 x \tan^2 x}{3}\\ &= \dfrac{1+0}{3} = \dfrac{1}{3} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{\cos \text{h}x - \cos x}{x \sin x}\)

Lets re-write the equation,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\cos \text{h}x - \cos x}{x^2 \dfrac{\sin x}{x}}\\ &= \lim_{x \to 0} \dfrac{\cos \text{h}x - \cos x}{x^2}, \text{ as } \dfrac{\sin x}{x}=1 \\ &= \lim_{x \to 0} \dfrac{\cos \text{h}x - \cos x}{x^2} \Longrightarrow \left(\dfrac{0}{0}\right) \end{split} \end{equation*}\]

Applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \dfrac{\sin \text{h}x + \sin x}{2x} & \Longrightarrow \left(\dfrac{0}{0}\right) \\ &= \dfrac{\cos \text{h}x + \cos x}{2} & \Longrightarrow \left(\dfrac{0}{0}\right) \\ &= \dfrac{2}{2} = 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{(e^{3x} -1)\tan^2 x}{x^3}\)

The expression can be written as,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{(e^{3x} -1)\left(\dfrac{\tan x}{x}\right)^2 \times x^2}{x^3}\\ \end{split} \end{equation*}\]

Using property of multiplication of two limits,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{(e^{3x} -1) x^2}{x^3} \times \lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^2 \\ &= \lim_{x \to 0} \dfrac{(e^{3x} -1) x^2}{x^3}, \text{ as } \lim_{x \to 0} \dfrac{\tan x}{x} =0 \\ &= \lim_{x \to 0} \dfrac{(e^{3x} -1)}{x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ \end{split} \end{equation*}\]

Applying L’Hospital rule,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{3 e^{3x}}{1}\\ &= 3 \end{split} \end{equation*}\]

9.1.2 Question 2

Evaluate the limits:

  1. \(\lim_{x \to 0} \dfrac{\log \sin x}{\cot x}\)

Form of \(\dfrac{\infty}{\infty}\), so,

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\dfrac{\cos x}{\sin x}}{-\csc^2 x}\\ &= \lim_{x \to 0} -\sin x \cos x\\ &= 0 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \dfrac{\log \tan x}{\log x}\)

This is a limit of form \(\dfrac{\infty}{\infty}\), so

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\dfrac{1}{\tan x}\times \sec^2 x}{\dfrac{1}{x}}\\ &= \lim_{x \to 0} \dfrac{x}{\sin x \cos x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{1}{\cos^2 x - \sin^2 x}\\ &= \dfrac{1}{1-0} = 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \log_{\tan x}\tan 2x\)

Properties of logarithms

  • If \(y = \log_b x\), then \(x = b^y\)
  • \(\log_b a = \dfrac{\log a}{\log b}\)

The given expression can then be written as,

\[\begin{equation*} \begin{split} \lim_{x \to 0} \log_{\tan x}\tan 2x &= \lim_{x \to 0} \dfrac{\log \tan 2x}{\log \tan x} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{2 \sec^2 2x}{\tan 2x}}{\dfrac{\sec^2 x}{\tan x}}\\ &= \lim_{x \to 0} \dfrac{2 \sin x \cos x}{\sin 2x \cos 2x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2(\cos^2 x - \sin^2 x)}{2(\cos^2 2x - \sin^2 2x)}\\ &= \dfrac{2}{2} = 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} x \log x\)

The limit is of form \(0 \times \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\log x}{\dfrac{1}{x}} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^2}}\\ &= \lim_{x \to 0} -x \\ &= 0 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} x \log \tan x\)

This is of form \(0 \times \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\log \tan x}{\dfrac{1}{x}} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right) \\ &= \lim_{x \to 0} \dfrac{\dfrac{\sec^2 x}{\tan x}}{-\dfrac{1}{x^2}}\\ &= \lim_{x \to 0} -\dfrac{x^2}{\sin x \cos x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} -\dfrac{2x}{\cos^2 x - \sin^2 x}\\ &= \dfrac{0}{1} = 0 \end{split} \end{equation*}\]

  1. \(\lim_{x \to a} (a-x) \tan \left(\dfrac{\pi x}{2a}\right)\)

This is of form \(0 \times \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to a} \dfrac{a-x}{\cot \dfrac{\pi x}{2a}} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to a} \dfrac{-1}{- \csc^2 \dfrac{\pi x}{2a} \times \dfrac{\pi}{2a}}\\ &= \lim_{x \to a} \sin^2 \dfrac{\pi x}{2a} \times \dfrac{2a}{\pi}\\ &= \dfrac{2a}{\pi} \end{split} \end{equation*}\]

  1. \(\lim_{x \to \infty} \dfrac{x^n}{e^x}\), \(n\) being a positive integer

This is of form \(\dfrac{\infty}{\infty}\),

\[\begin{equation*} \begin{split} \lim_{x \to \infty} \dfrac{x^n}{e^x} &= \lim_{x \to \infty} \dfrac{nx^{n-1}}{e^x} \\ &= \lim_{x \to \infty} \dfrac{n(n-1)x^{n-2}}{e^x} = \ldots = \lim_{x \to \infty} \dfrac{n!}{e^x}\\ &= 0 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} x \log \sin^2 x\)

The expression can be written as, \(\lim_{x \to 0} 2x \log \sin x\)

This is of form \(0 \times \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{2 \log \sin x}{\dfrac{1}{x}} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to 0} \dfrac{2 \cot x}{-1 x^{-2}}\\ &= \lim_{x \to 0} -\dfrac{2x^2}{\tan x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} -\dfrac{4x}{\sec^2 x}\\ &= \dfrac{0}{1} = 0 \end{split} \end{equation*}\]

9.1.3 Question 3

Evaluate the limits:

  1. \(\lim_{x \to 0} \left(\dfrac{1}{x^2}- \dfrac{1}{\sin^2 x}\right)\) [TU 2063, 64]

This is of form \(\infty - \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{\sin^2 x - x^2}{x^2 \sin^2 x}\\ &= \lim_{x \to 0} \dfrac{\sin^2 x - x^2}{x^4 \times \left(\dfrac{\sin x}{x}\right)^2}\\ &= \lim_{x \to 0} \dfrac{\sin^2 x - x^2}{x^4}, \lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)=1\\ &= \lim_{x \to 0} \dfrac{\sin^2 x - x^2}{x^4} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2\sin x \cos x - 2x}{4x^3} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2(\cos^2 x -\sin^2 x) -2}{12x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2\cos 2x -2}{12x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-4\sin 2x}{24x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-8\cos 2x}{24}\\ &= -\dfrac{1}{3} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \left(\dfrac{1}{x} - \dfrac{1}{e^x - 1}\right)\)

This is of form \(\infty - \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{e^x - x -1}{x(e^x -1)}\\ &= \lim_{x \to 0} \dfrac{e^x - x -1}{x^2 \times \dfrac{e^x -1}{x}}\\ &= \lim_{x \to 0} \dfrac{e^x - x- 1}{x^2} & \Longrightarrow \lim_{x \to 0} \dfrac{e^x -1}{x} = 1\\ &= \lim_{x \to 0} \dfrac{e^x - x- 1}{x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{e^x -1}{2x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{e^x}{2}\\ &= \dfrac{1}{2} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \left(\dfrac{1}{x^2}- \cot^2 x\right)\)

This is of form \(\infty - \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{1}{x^2} - \dfrac{\cos^2 x}{\sin^2 x}\\ &= \frac{\sin^2 x - x^2 \cos^2 x}{x^2 \sin^2 x}\\ \end{split} \end{equation*}\]

After we put the expansion of \(\sin x\) and \(\cos x\), we get,

\[\begin{equation*} \begin{split} &= \dfrac{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots\right)^2 - x^2\left(1 -\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots\right)^2}{x^2 \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots\right)^2}\\ &= \lim_{x \to 0} \dfrac{\dfrac{2}{3}x^4 + \ldots}{x^4 + \ldots} \\ &= \dfrac{2}{3} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \left(\dfrac{1}{x} - \dfrac{1}{x^2}\log (1+x)\right)\)

This is of form \(\infty - \infty\).

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{x - \log(1+x)}{x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{1 - \frac{1}{1+x}}{2x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{(1+x)^{-2}}{2}\\ &= \dfrac{1}{2} \end{split} \end{equation*}\]

9.1.4 Question 4

Evaluate the limits:

  1. \(\lim_{x \to 0} x^x\)

This is of form \(0^0\).

\[\begin{equation*} \begin{split} y &= x^x \\ \log y &= \log x^x\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} x \log x & \Longrightarrow 0 \times \infty \\ &= \lim_{x \to 0} \dfrac{\log x}{\dfrac{1}{x}} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ \end{split} \end{equation*}\]

Applying L’Hospital rule,

\[\begin{equation*} \begin{split} \lim_{x \to 0} \log y &= \lim_{x \to 0} \dfrac{\dfrac{1}{x}}{-x^{-2}}\\ &= \lim_{x \to 0} - x \\ \lim_{x \to 0} \log y &= 0\\ \log \left(\lim_{x \to 0} y\right) &= 0\\ \lim_{x \to 0} y &= e^0\\ \lim_{x \to 0} x^x &= 1\\ \end{split} \end{equation*}\]

  1. \(\lim_{x \to \pi/2} (\sin x)^{\tan x}\)

\[\begin{equation*} \begin{split} \lim_{x \to 0} \log y &= \log \left(\lim_{x \to 0} y\right) \end{split} \end{equation*}\]

The limit is of form \(1^{\infty}\).

\[\begin{equation*} \begin{split} y &= (\sin x)^{\tan x} \\ \log y &= \log (\sin x)^{\tan x}\\ \log y &= \tan x \log \sin x\\ \lim_{x \to \pi/2} \log y &= \lim_{x \to \pi/2} \dfrac{\log \sin x}{\cot x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to \pi/2} \dfrac{\dfrac{\cos x}{\sin x}}{-\csc^2 x}\\ &= \lim_{x \to \pi/2} -\sin x \cos x\\ \lim_{x \to \pi/2} \log y &= 0\\ \log \left(\lim_{x \to \pi/2} y\right) &= 0\\ \lim_{x \to \pi/2} y &= e^0\\ \lim_{x \to \pi/2} (\sin x)^{\tan x} &= 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} (\cot x)^{\sin 2x}\)

It is of form \({\infty}^0\).

\[\begin{equation*} \begin{split} y &= (\cot x)^{\sin 2x}\\ \log y &= \sin 2x \log \cot x \\ \lim_{x \to 0} \log y &= \lim_{x \to 0} \sin 2x \log \cot x & \Longrightarrow 0 \times \infty\\ &= \lim_{x \to 0} \dfrac{\log \cot 2x}{\csc 2x} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{-2\csc^2 2x}{\cot 2x}}{-2 \csc 2x \cot 2x}\\ &= \lim_{x \to 0} \dfrac{\csc 2x}{\cot 2x} \times \dfrac{1}{\cot 2x}\\ &= \lim_{x \to 0} \dfrac{\tan 2x}{\cos 2x} = \dfrac{0}{1}\\ \lim_{x \to 0} \log y &= 0\\ \lim_{x \to 0} y &= e^0\\ \lim_{x \to 0} (\cot x)^{\sin 2x} &= 1\\ \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} (\cos x)^{\cot^2 x}\) [TU 2058]

It is of form \(1^{\infty}\).

\[\begin{equation*} \begin{split} y &= (\cos x)^{\cot^2 x}\\ \log y &= \cot^2 x \log \cos x\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} \cot^2 x \log \cos x & \Longrightarrow \infty \times 0\\ &= \lim_{x \to 0} \dfrac{\log \cos x}{\tan^2 x} \\ &= \lim_{x \to 0} \dfrac{\log \cos x}{x^2 \left(\dfrac{\tan x}{x}\right)^2}\\ &= \lim_{x \to 0} \dfrac{\log \cos x}{x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-\tan x}{2x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-\sec^2 x}{2}\\ \lim_{x \to 0} \log y &= -\dfrac{1}{2}\\ \lim_{x \to 0} (\cos x)^{\cot^2 x} &= e^{-1/2} \end{split} \end{equation*}\]

  1. \(\lim_{x \to \pi} (\sin x)^{\tan x}\)

It is of form \(0^{\infty}\).

\[\begin{equation*} \begin{split} y &= (\sin x)^{\tan x} \\ \log y &= \tan x \log \sin x \\ \lim_{x \to \pi} \log y &= \lim_{x \to \pi} \tan x \log \sin x & \Longrightarrow 0 \times \infty\\ &= \lim_{x \to \pi} \dfrac{\log \sin x}{\cot x} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to \pi} \dfrac{\cot x}{-\csc^2 x}\\ &= \lim_{x \to \pi} -\sin x \cos x \\ \lim_{x \to \pi} \log y &= 0\\ \lim_{x \to \pi} y &= e^0\\ \lim_{x \to \pi} (\sin x)^{\tan x} &= 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} (\cot^2 x)^{\sin x}\) [TU 2054, 2060, 2066]

The limit is of form \({\infty}^0\).

\[\begin{equation*} \begin{split} y &= (\cot^2 x)^{\sin x}\\ \log y &= \sin x \log (\cot^2 x)\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} \sin x \log (\cot^2 x) & \Longrightarrow 0 \times \infty \\ &= \lim_{x \to 0} \dfrac{\log (\cot^2 x)}{\csc x} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to 0} \dfrac{\frac{1}{\cot^2 x} \times 2 \cot x \times (-\csc^2 x)}{-\csc x \cot x}\\ &= \lim_{x \to 0} \dfrac{2 \sin x}{\cos^2 x}\\ \lim_{x \to 0} \log y &= 0\\ \lim_{x \to 0} y &= e^0\\ \lim_{x \to 0} (\cot^2 x)^{\sin x} &= 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \left(\dfrac{1}{x^2}\right)^{\tan x}\)

The limit will be of form \({\infty}^0\).

\[\begin{equation*} \begin{split} y &= \left(\dfrac{1}{x^2}\right)^{\tan x} \\ \log y &= \tan x \log \left(\dfrac{1}{x^2}\right)\\ &= \tan x(\log 1 - \log x^2)\\ &= -2 \log x \tan x\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} -2 \log x \tan x & \Longrightarrow \infty \times 0\\ &= \lim_{x \to 0} -\dfrac{2 \log x}{\cot x} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to 0} -\dfrac{\frac{2}{x}}{-\csc^2 x}\\ &= \lim_{x \to 0} \dfrac{2\sin^2 x}{x}\\ &= \lim_{x \to 0} 2x \left(\dfrac{\sin x}{x}\right)^2 \\ &= \lim_{x \to 0} 2x\\ \lim_{x \to 0} \log y &= 0\\ \lim_{x \to 0} y &= e^0\\ \lim_{x \to 0} \left(\dfrac{1}{x^2}\right)^{\tan x} &= 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^{1/x}\) [TU 2057, 2062, 2063]

Let,

\[\begin{equation*} \begin{split} y &= \left(\dfrac{\tan x}{x}\right)^{1/x}\\ \log y &= \dfrac{1}{x} \log \left(\dfrac{\tan x}{x}\right)\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} \dfrac{1}{x} \log \left(\dfrac{\tan x}{x}\right)\\ &= \lim_{x \to 0} \dfrac{\log \left(\dfrac{\tan x}{x}\right)}{x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{1}{\dfrac{\tan x}{x}}\times \dfrac{\mathrm{d}}{\mathrm{dx}}\left(\dfrac{\tan x}{x}\right)}{1} & \Longrightarrow \lim_{x \to 0} \dfrac{\tan x}{x} = 1\\ &= \lim_{x \to 0} \dfrac{1}{1}\times \dfrac{\mathrm{d}}{\mathrm{dx}}\left(\dfrac{\tan x}{x}\right) \\ &= \lim_{x \to 0} \dfrac{x \sec^2 x - \tan x}{x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2x \sec^2 x \tan x +\sec^2 x - \sec^2 x}{2x}\\ &= \lim_{x \to 0} \sec^2 x \tan x \\ \lim_{x \to 0} \log y &= 0\\ \lim_{x \to 0} y &= e^0\\ \lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^{1/x} &= 1 \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)^{1/x^2}\)

This is of form \(1^{\infty}\).

\[\begin{equation*} \begin{split} y &= \left(\dfrac{\sin x}{x}\right)^{1/x^2} \\ \log y &= \dfrac{1}{x^2} \log \left(\dfrac{\sin x}{x}\right)\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} \dfrac{\log \left(\dfrac{\sin x}{x}\right)}{x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{1}{\dfrac{\sin x}{x}} \times \dfrac{\mathrm{d}}{\mathrm{dx}} \left(\dfrac{\sin x}{x}\right)}{2x}\\ &= \lim_{x \to 0} \dfrac{1 \times \dfrac{x \cos x - \sin x}{x^2}}{2x}\\ &= \lim_{x \to 0} \dfrac{x \cos x - \sin x}{2x^3} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-x \sin x + \cos x - \cos x}{6x^2}\\ &= \lim_{x \to 0} \dfrac{-\sin x}{6x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-\cos x}{6}\\ \lim_{x \to 0} \log y &= -\dfrac{1}{6}\\ \lim_{x \to 0} y &= e^{-1/6}\\ \lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)^{1/x^2} &= e^{-1/6}\\ \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} (\cot x)^{1/\log x}\)

This limit is of form \({\infty}^0\).

\[\begin{equation*} \begin{split} y &= (\cot x)^{1/\log x} \\ \log y &= \dfrac{1}{\log x} \log (\cot x)\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} \dfrac{\log (\cot x)}{\log x} & \Longrightarrow \left(\dfrac{\infty}{\infty}\right)\\ &= \lim_{x \to 0} \dfrac{\dfrac{1}{\cot x}\times (-\csc^2 x)}{\dfrac{1}{x}}\\ &= \lim_{x \to 0} -\dfrac{x}{\sin x \cos x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} -\dfrac{1}{\cos^2 x - \sin^2 x}\\ &= -1\\ \lim_{x \to 0} \log y &= -1\\ \lim_{x \to 0} y &= e^{-1}\\ \lim_{x \to 0} (\cot x)^{1/\log x} &= e^{-1} \end{split} \end{equation*}\]

  1. \(\lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^{1/x^2}\)

The limit of this expression is \(1^{\infty}\).

\[\begin{equation*} \begin{split} y &= \left(\dfrac{\tan x}{x}\right)^{1/x^2} \\ \log y &= \dfrac{1}{x^2} \log \left(\dfrac{\tan x}{x}\right)\\ \lim_{x \to 0} \log y &= \lim_{x \to 0} \dfrac{\log \left(\dfrac{\tan x}{x}\right)}{x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{\frac{1}{\frac{\tan x}{x}}\times \dfrac{\mathrm{d}}{\mathrm{dx}}\left(\dfrac{\tan x}{x}\right)}{2x} & \Longrightarrow \lim_{x \to 0} \dfrac{\tan x}{x} = 1\\ &= \lim_{x \to 0} \dfrac{x \sec^2 x - \tan x}{2x\times x^2} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2x \sec^2 x \tan x + \sec^2 x - \sec^2 x}{6x^2}\\ &= \lim_{x \to 0} \dfrac{\sec^2 x \tan x}{3x} & \Longrightarrow \lim_{x \to 0} \sec^2 x = 1 \\ &= \lim_{x \to 0} \dfrac{1 \times \tan x}{3x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{\sec^2 x}{3}\\ \lim_{x \to 0} \log y &= \dfrac{1}{3}\\ \lim_{x \to 0} y &= e^{1/3}\\ \lim_{x \to 0} \left(\dfrac{\tan x}{x}\right)^{1/x^2} &= e^{1/3}\\ \end{split} \end{equation*}\]