Chapter 10 Indeterminate Forms-II

10.1 Exercise 5

10.1.1 Question 5

Evaluate the following limits

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

This is of form $1^{\infty}$ .

$\begin{aligned} y & = x^{1 / (1 - x)} \\ \log y & = \frac{1}{1 - x} \log x \\ lim_{x \to 1} \log y & = lim_{x \to 1} \frac{\log x}{1 - x} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 1} \frac{\frac{1}{x}}{- 1} \\ = - 1 \\ lim_{x \to 1} \log y & = - 1 \\ lim_{x \to 1} y & = e^{- 1} \\ lim_{x \to 1} x^{1 / (1 - x)} & = e^{- 1} \end{aligned}$

$lim_{x \to \infty} x^{1 / x}$

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

$\begin{aligned} y & = x^{1 / x} \\ \log y & = \frac{1}{x} \log x \\ lim_{x \to \infty} \log y & = lim_{x \to \infty} \frac{\log x}{x} & ⟹ (\frac{\infty}{\infty}) \\ = lim_{x \to \infty} \frac{\frac{1}{x}}{1} \\ = lim_{x \to \infty} \frac{1}{x} \\ = 0 \\ lim_{x \to \infty} \log y & = 0 \\ lim_{x \to \infty} y & = e^{0} \\ lim_{x \to \infty} x^{1 / x} & = 1 \end{aligned}$

$lim_{x \to 0} \frac{(1 + x)^{1 / x} - e}{x}$

The given expression can be written as,

$\begin{aligned} \frac{(1 + x)^{1 / x} - e}{x} & = \frac{e^{\frac{1}{x} \log (1 + x)} - e}{x} \end{aligned}$

So,

$\begin{aligned} = lim_{x \to 0} \frac{e^{\frac{1}{x} \log (1 + x)} - e}{x} \\ = lim_{x \to 0} \frac{e^{\frac{1}{x} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} - \dots)} - e}{x} \\ = lim_{x \to 0} \frac{e^{(1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots)} - e}{x} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{e^{(1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots)} \times (0 - \frac{1}{2} + \frac{2 x}{3} - \dots) - 0}{1} \\ = e \times (- \frac{1}{2}) \\ = - \frac{e}{2} \end{aligned}$

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

$lim_{x \to 0} \frac{x e^{x} - (1 + x) \log (1 + x)}{x^{2}}$

$\begin{aligned} = lim_{x \to 0} \frac{x e^{x} - (1 + x) \log (1 + x)}{x^{2}} & ⟹ (\frac{0}{0}) \\ = \frac{x e^{x} + e^{x} - (\frac{1 + x}{1 + x} + \log (1 + x))}{2 x} \\ = \frac{x e^{x} + e^{x} - 1 - \log (1 + x)}{2 x} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{x e^{x} + e^{x} + e^{x} - 0 - \frac{1}{1 + x}}{2} \\ = \frac{0 + 1 + 1 - 1}{2} \\ = \frac{1}{2} \end{aligned}$

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

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

$\begin{aligned} y & = (\cos x)^{1 / x^{2}} \\ \log y & = \frac{1}{x^{2}} \log (\cos x) \\ lim_{x \to 0} \log y & = lim_{x \to 0} \frac{\log (\cos x)}{x^{2}} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{- \tan x}{2 x} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} - \frac{\sec^{2} x}{2} \\ lim_{x \to 0} \log y & = - \frac{1}{2} \\ lim_{x \to 0} y & = e^{- 1 / 2} \\ lim_{x \to 0} (\cos x)^{1 / x^{2}} & = e^{- 1 / 2} \end{aligned}$

$lim_{x \to a} {(2 - \frac{x}{a})}^{\tan \frac{π x}{2 a}}$

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

$\begin{aligned} y & = {(2 - \frac{x}{a})}^{\tan \frac{π x}{2 a}} \\ lim_{x \to a} \log y & = lim_{x \to a} \tan \frac{π x}{2 a} \log (2 - \frac{x}{a}) & ⟹ \infty \times 0 \\ = lim_{x \to a} \frac{\log (2 - \frac{x}{a})}{\cot \frac{π x}{2 a}} & ⟹ (\frac{0}{0}) \\ = lim_{x \to a} \frac{\frac{1}{2 - \frac{x}{a}} \times (- \frac{1}{a})}{- \csc^{2} \frac{π x}{2 a} \times \frac{π}{2 a}} \\ = \frac{- \frac{1}{a}}{- \frac{π}{2 a}} \\ lim_{x \to a} \log y & = \frac{2}{π} \\ lim_{x \to a} y & = e^{\frac{2}{π}} \\ lim_{x \to a} {(2 - \frac{x}{a})}^{\tan \frac{π x}{2 a}} & = e^{\frac{2}{π}} \end{aligned}$

10.1.2 Question 6

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

The given expression takes the form $\frac{0}{0}$ as $x \to 0$ for any value of $a$ .

$\begin{aligned} = lim_{x \to 0} \frac{a \sin x - \sin 2 x}{\tan^{3} x} \\ = lim_{x \to 0} \frac{a \sin x - \sin 2 x}{x^{3} {(\frac{\tan x}{x})}^{3}} \\ = lim_{x \to 0} \frac{a \sin x - \sin 2 x}{x^{3}} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{a \cos x - 2 \cos 2 x}{3 x^{2}} \end{aligned}$

This takes the form $\frac{a - 2}{0}$ as $x \to 0$ . In order that the limit be finite, the form should be $\frac{0}{0}$ . So,

$\begin{aligned} a - 2 & = 0 \\ a & = 2 \end{aligned}$

Then the limit,

$\begin{aligned} = lim_{x \to 0} \frac{2 \cos x - 2 \cos 2 x}{3 x^{2}} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{- 2 \sin x + 4 \sin 2 x}{6 x} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{- 2 \cos x + 8 \cos 2 x}{6} \\ = \frac{- 2 + 8}{6} \\ = 1 \end{aligned}$

Determine the values of $a$ and $b$ , so that:

$\begin{aligned} lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{3}} & = 1 \end{aligned}$

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

$\begin{aligned} = lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{3}} & ⟹ (\frac{0}{0}) \\ (10.1) & = lim_{x \to 0} \frac{(1 + a \cos x) - a x \sin x - b \cos x}{3 x^{2}} \end{aligned}$

This takes the form $\frac{1 + a - 0 - b}{0}$ as $x \to 0$ . In order that the limit be finite i.e. $1$ , the form should be $\frac{0}{0}$ . So,

$\begin{aligned} (10.2) & a - b + 1 & = 0 \end{aligned}$

Continuing with (10.1),

$\begin{aligned} = lim_{x \to 0} \frac{- a \sin x - a (x \cos x + \sin x) + b \sin x}{6 x} \\ = lim_{x \to 0} \frac{- 2 a \sin x - a x \cos x + b \sin x}{6 x} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{- 2 a \cos x - a (\cos x - x \sin x) + b \cos x}{6} \\ = \frac{- 2 a - a + b}{6} \end{aligned}$

The limit is given as $1$ . So,

$\begin{aligned} \frac{- 2 a - a + b}{6} & = 1 \\ (10.3) & - 3 a + b & = 6 \end{aligned}$

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

$\begin{aligned} a & = - \frac{5}{2} \\ b & = - \frac{3}{2} \end{aligned}$

Determine the values of $a$ , $b$ , $c$ , so that:

$\begin{aligned} lim_{x \to 0} \frac{(a + b \cos x) x - c \sin x}{x^{5}} & = 1 \end{aligned}$

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

$\begin{aligned} = lim_{x \to 0} \frac{(a + b \cos x) x - c \sin x}{x^{5}} & ⟹ (\frac{0}{0}) \\ (10.4) & = lim_{x \to 0} \frac{- b x \sin x + a + b \cos x - c \cos x}{5 x^{4}} \end{aligned}$

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

$\begin{aligned} (10.5) & a + b - c & = 0 \end{aligned}$

Continuing from (10.4),

$\begin{aligned} = lim_{x \to 0} \frac{- b (x \cos x + \sin x) + 0 - b \sin x + c \sin x}{20 x^{3}} & ⟹ (\frac{0}{0}) \\ (10.6) & = lim_{x \to 0} \frac{- b (\cos x - x \sin x + \cos x) - b \cos x + c \cos x}{60 x^{2}} \end{aligned}$

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

$\begin{aligned} (10.7) & 3 b - c & = 0 \end{aligned}$

Continuing with (10.6),

$\begin{aligned} = lim_{x \to 0} \frac{- b (- 2 \sin x - x \cos x - \sin x) + b \sin x - c \sin x}{120 x} & ⟹ (\frac{0}{0}) \\ = lim_{x \to 0} \frac{2 b \cos x + b (\cos x - x \sin x + \cos x) + b \cos x - c \cos x}{120} \end{aligned}$

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

$\begin{aligned} \frac{2 b + b (1 - 0 + 1) + b - c}{120} & = 1 \\ (10.8) & 5 b - c & = 120 \end{aligned}$

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

$\begin{aligned} a & = 120 \\ b & = 60 \\ c & = 180 \end{aligned}$