Chapter 10 Indeterminate Forms-II

10.1 Exercise 5

10.1.1 Question 5

Evaluate the following limits

  1. \(\lim_{x \to 1} x^{1/(1-x)}\) [TU 2056]

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

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

  1. \(\lim_{x \to \infty} x^{1/x}\)

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

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

  1. \(\lim_{x \to 0} \dfrac{(1+x)^{1/x} - e}{x}\)

The given expression can be written as,

\[\begin{equation*} \begin{split} \dfrac{(1+x)^{1/x} - e}{x} &= \dfrac{e^{\frac{1}{x}\log (1+x)} - e}{x} \end{split} \end{equation*}\]

So,

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

Derivative of \(e^{\left(1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \ldots\right)}\) is \(e^{\left(1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \ldots\right)} \times \left(0 -\frac{1}{2} + \frac{2x}{3} - \ldots\right)\).

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

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

  1. \(\lim_{x \to 0} (\cos x)^{1/x^2}\)

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

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

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

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

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

10.1.2 Question 6

  1. If \(\lim_{x \to 0} \dfrac{a\sin x - \sin 2x}{\tan^3 x}\) is finite, find the value of \(a\) and the limit.

The given expression takes the form \(\frac{0}{0}\) as \(x \rightarrow 0\) for any value of \(a\).

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

This takes the form \(\dfrac{a -2}{0}\) as \(x \rightarrow 0\). In order that the limit be finite, the form should be \(\dfrac{0}{0}\). So,

\[\begin{equation*} \begin{split} a -2 &= 0\\ a &= 2 \end{split} \end{equation*}\]

Then the limit,

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

  1. Determine the values of \(a\) and \(b\), so that:

\[\begin{equation*} \begin{split} \lim_{x \to 0} \dfrac{x(1+ a \cos x) - b \sin x}{x^3} &=1 \end{split} \end{equation*}\]

The given expression takes the form \(\frac{0}{0}\) as \(x \rightarrow 0\) for any value of \(a\) and \(b\).

\[\begin{align} &= \lim_{x \to 0} \dfrac{x(1+ a \cos x) - b \sin x}{x^3} & \Longrightarrow \left(\dfrac{0}{0}\right) \notag \\ &= \lim_{x \to 0} \dfrac{(1 + a\cos x) - ax \sin x - b\cos x}{3x^2} \tag{10.1} \end{align}\]

This takes the form \(\dfrac{1 + a-0-b}{0}\) as \(x \rightarrow 0\). In order that the limit be finite i.e. \(1\), the form should be \(\dfrac{0}{0}\). So,

\[\begin{align} a- b + 1 &= 0 \tag{10.2} \end{align}\]

Continuing with (10.1),

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{-a \sin x - a(x\cos x + \sin x) + b\sin x}{6x}\\ &= \lim_{x \to 0} \dfrac{-2a \sin x - ax \cos x + b \sin x}{6x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{-2a \cos x -a(\cos x - x\sin x) + b\cos x}{6}\\ &= \dfrac{-2a -a + b}{6} \end{split} \end{equation*}\]

The limit is given as \(1\). So,

\[\begin{align} \dfrac{-2a -a + b}{6} &= 1 \notag \\ -3a + b &= 6 \tag{10.3} \end{align}\]

Solving (10.2) and (10.3), we get,

\[\begin{equation*} \begin{split} a &= -\dfrac{5}{2}\\ b &= -\dfrac{3}{2} \end{split} \end{equation*}\]

  1. Determine the values of \(a\), \(b\), \(c\), so that:

\[\begin{equation*} \begin{split} \lim_{x \to 0} \dfrac{(a + b \cos x)x - c \sin x}{x^5} &= 1 \end{split} \end{equation*}\]

The LHS of given expression takes the form \(\frac{0}{0}\) as \(x \rightarrow 0\) for any value of \(a\), \(b\) and \(c\).

\[\begin{align} &= \lim_{x \to 0} \dfrac{(a + b \cos x)x - c \sin x}{x^5} & \Longrightarrow \left(\dfrac{0}{0}\right) \notag\\ &= \lim_{x \to 0} \dfrac{-bx \sin x + a + b\cos x - c \cos x}{5x^4} \tag{10.4} \end{align}\]

This takes the form \(\dfrac{a + b -c}{0}\) as \(x \rightarrow 0\). In order that the limit be finite i.e. \(1\), the form should be \(\dfrac{0}{0}\). So,

\[\begin{align} a + b -c &= 0 \tag{10.5} \end{align}\]

Continuing from (10.4),

\[\begin{align} &= \lim_{x \to 0} \dfrac{-b(x \cos x + \sin x) + 0 - b\sin x + c \sin x}{20x^3} & \Longrightarrow \left(\dfrac{0}{0}\right) \notag \\ &= \lim_{x \to 0} \dfrac{-b(\cos x - x \sin x + \cos x) - b\cos x + c\cos x}{60x^2} \tag{10.6} \end{align}\]

This takes the form \(\dfrac{-2b - b + c}{0}\) as \(x \rightarrow 0\). In order that the limit be finite i.e. \(1\), the form should be \(\dfrac{0}{0}\). So,

\[\begin{align} 3b -c &= 0 \tag{10.7} \end{align}\]

Continuing with (10.6),

\[\begin{equation*} \begin{split} &= \lim_{x \to 0} \dfrac{-b(-2 \sin x - x\cos x - \sin x) + b\sin x - c\sin x}{120x} & \Longrightarrow \left(\dfrac{0}{0}\right)\\ &= \lim_{x \to 0} \dfrac{2b\cos x + b(\cos x -x\sin x + \cos x) + b\cos x - c\cos x}{120} \end{split} \end{equation*}\]

This takes the form \(\dfrac{2b + b (1-0 +1) + b - c}{120}\) as \(x \rightarrow 0\). Also given the limit of expression is \(1\). Equating two, we have,

\[\begin{align} \dfrac{2b + b (1-0 +1) + b - c}{120} &= 1 \notag \\ 5b -c &= 120 \tag{10.8} \end{align}\]

Solving (10.5), (10.7) and (10.8), we get,

\[\begin{equation*} \begin{split} a &= 120\\ b &= 60 \\ c &= 180 \end{split} \end{equation*}\]