Limits Indeterminate Form

Evaluate: $\lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right]$

$\begin{aligned} \lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(x \cos x - \sin x\right)}{\dfrac{d}{dx} \left(x^2 \sin x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x - x \sin x - \cos x}{2x \sin x + x^2 \cos x}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- x \sin x}{x \left(2 \sin x + x \cos x\right)}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- \sin x}{2 \sin x + x \cos x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(- \sin x\right)}{\dfrac{d}{dx} \left(2 \sin x + \cos x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x}{2 \cos x + \sin x}\right] \\\\ & = \dfrac{1}{2 + 0} \\\\ & = \dfrac{1}{2} \end{aligned}$

Limits Indeterminate Form

Evaluate $\;$ $\lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y}$

$\begin{aligned} \lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[y - \tan^{-1}y\right]}{\dfrac{d}{dy} \left[y - \sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{1 - \dfrac{1}{1 + y^2}}{1 - \cos y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[1 - \dfrac{1}{1 + y^2}\right]}{\dfrac{d}{dy} \left[1 - \cos y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{1}{\left(1 + y^2\right)^2} \times 2y}{\sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[\dfrac{y}{\left(1 + y^2\right)^2}\right]}{\dfrac{d}{dy} \left[\sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{\left(1 + y^2\right)^2 \times 1 - y \times 2 \left(1 + y^2\right) \times 2y}{\left(1 + y^2\right)^4}}{\cos y} \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\left(1 + y^2\right)^2 - 4 y^2 \left(1 + y^2\right)}{\left(1 + y^2\right)^4 \cos y} \\\\ & = 2 \times \left[\dfrac{1 - 0}{1 \times 1}\right] \\\\ & = 2 \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 1} \; \left(\dfrac{1}{\log x}-\dfrac{1}{x-1}\right)$

$\begin{aligned} & \lim\limits_{x \to 1} \; \left(\dfrac{1}{\log x}-\dfrac{1}{x-1}\right) & \left(\infty - \infty \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{x-1-\log x}{\left(x-1\right)\log x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{\dfrac{d}{dx}\left[x-1-\log x\right]}{\dfrac{d}{dx}\left[\left(x-1\right)\log x\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{1-0-\dfrac{1}{x}}{\dfrac{x-1}{x}+\log x} & \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{x-1}{x-1+\log x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{\dfrac{d}{dx}\left(x-1\right)}{\dfrac{d}{dx}\left(x-1+\log x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{1}{1+\dfrac{1}{x}} & \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{x}{x+1} = \dfrac{1}{1+1} = \dfrac{1}{2} & \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{\left(\tan^{-1}x\right)^2}{\log \left(1+x^2\right)}$

$\begin{aligned} &\lim\limits_{x \to 0} \; \dfrac{\left(\tan^{-1}x\right)^2}{\log \left(1+x^2\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\tan^{-1}x\right)^2}{\dfrac{d}{dx}\left[\log \left(1+x^2\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2 \tan^{-1}x / \left(1+x^2\right)}{2x / \left(1+x^2\right)} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\tan^{-1}x}{x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\tan^{-1}x\right)}{\dfrac{d}{dx}\left(x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{1/1+x^2}{1} & \\ & & \\ & = \dfrac{1}{1+0} = 1 & \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{e^x \sin x - x -x^2}{x^2 + x \log \left(1-x\right)}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{e^x \sin x -x -x^2}{x^2 + x \log \left(1-x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{e^x \sin x -x - x^2\right\}}{\dfrac{d}{dx}\left\{x^2 + x \log \left(1-x\right)\right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x \cos x + e^x \sin x -1 - 2x}{2x-\dfrac{x}{1-x}+\log \left(1-x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{e^x \cos x + e^x \sin x -1 -2x\right\}}{\dfrac{d}{dx}\left\{2x - \dfrac{x}{1-x}+ \log \left(1-x\right) \right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x \cos x -e^x \sin x + e^x \sin x + e^x \cos x -2}{2 - \dfrac{1-x+x}{\left(1-x\right)^2}- \dfrac{1}{1-x}} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2e^x \cos x -2}{2 - \dfrac{1}{\left(1-x\right)^2}-\dfrac{1}{1-x}} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{2e^x \cos x -2\right\}}{\dfrac{d}{dx}\left\{2-\dfrac{1}{\left(1-x\right)^2}-\dfrac{1}{1-x}\right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2\left(e^x \cos x -e^x \sin x\right)}{\dfrac{-2}{\left(1-x\right)^3}- \dfrac{1}{\left(1-x\right)^2} } & \\ & & \\ & = \dfrac{2\times \left(1-0\right)}{-2-1} & \\ & & \\ & = - \dfrac{2}{3} \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{x^2 e^x}{\tan^2 x }$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{x^2 e^x}{\tan^2 x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(x^2 e^x\right)}{\dfrac{d}{dx}\tan^2 x} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{x^2 e^x + 2x e^x}{2 \tan x \sec^2 x} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{xe^x \left(x+2\right)}{2 \tan x \sec^2 x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left[xe^x \left(x+2\right)\right]}{\dfrac{d}{dx}\left(2 \tan x \sec^2 x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{xe^x + \left(x+2\right)\left(xe^x + e^x\right)}{2 \left(\tan x \times 2\sec x \times \sec x \tan x + \sec^2 x \times \sec^2 x\right) } & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{4xe^x + x^2 e^x + 2e^x}{4 \tan^2 x \sec^2 x + 2 \sec^4 x} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x\left(x^2+4x+2\right)}{4\sec^2 x \left(\sec^2 x -1\right)+2\sec^4 x} & \\ & & \\ & = \dfrac{1\left(0+0+2\right)}{4 \times 1 \times \left(1-1\right)+ 2 \times 1} = \dfrac{2}{2} = 1 \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 1^{-}} \; \log \left(1-x\right) \cot \left(\dfrac{\pi x}{2}\right)$

$\begin{aligned} & \lim\limits_{x \to 1^{-}} \; \log \left(1-x\right) \cot \left(\dfrac{\pi x}{2}\right) & \left(\infty-0 \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\log \left(1-x\right)}{\tan \left(\dfrac{\pi x}{2}\right)} & \left(\dfrac{\infty}{\infty} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{d}{dx}\left[\log \left(1-x\right)\right]}{\dfrac{d}{dx}\left[\tan \left(\dfrac{\pi x}{2}\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{-1}{1-x}}{\dfrac{\pi}{2}\sec^2 \left(\dfrac{\pi x}{2}\right)} & \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{\cos^2 \left(\dfrac{\pi x}{2}\right)}{x-1} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{d}{dx}\left[\cos^2 \left(\dfrac{\pi x}{2}\right)\right]}{\dfrac{d}{dx}\left(x-1\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{-2 \cos \left(\dfrac{\pi x}{2}\right) \times \sin \left(\dfrac{\pi x}{2}\right) \times \dfrac{\pi}{2}}{1} & \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; - \sin \left(\dfrac{2 \pi x}{2}\right) & \left[\text{Note: }\sin 2 \theta = 2 \sin \theta \cos \theta\right] \\ & & \\ & = - \lim\limits_{x \to 1^{-}} \; \sin \left(\pi x\right) & \\ & & \\ & = 0 \end{aligned}$

Limits Indeterminate Form

If $\lim\limits_{x \to 0} \; \dfrac{\sin 2x + k \sin x}{x^3}$ is finite, find k and the limit.

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\sin 2x + k \sin x}{x^3} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin 2x + k \sin x\right)}{\dfrac{d}{dx}\left(x^3\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2 \cos 2x + k \cos x}{3x^2} & \cdots (1) \end{aligned}$

Equation (1) will be of $\dfrac{0}{0}$ form if $\;\;$ $2 \cos 0 + \cos 0 = 0$

i.e. $2+k = 0$ $\implies$ $k = -2$

$\therefore$ $\;$ Equation (1) becomes

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{2 \cos 2x - 2 \cos x}{3x^2} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\cos 2x - \cos x\right)}{\dfrac{d}{dx}\left(x^2\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{-2\sin 2x + \sin x}{2x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(-2\sin 2x + \sin x\right)}{\dfrac{d}{dx}\left(2x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{-4\cos 2x + \cos x}{2} & \\ & & \\ & = \dfrac{1}{3} \times \left(-4\cos 0 + \cos 0\right) & \\ & & \\ & = \dfrac{1}{3} \times \left(-4+1\right) = -1 \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to \frac{\pi}{4}} \; \left(\tan x\right)^{\tan 2x}$

Let $f\left(x\right)=\left(\tan x\right)^{\tan 2x}$

Then, $\log \left[f\left(x\right)\right] = \tan 2x \log \left(\tan x\right)$

$\begin{aligned} \therefore \; \lim\limits_{x \to \frac{\pi}{4}} \; \left\{\log \left[f\left(x\right)\right] \right\} & = \lim\limits_{x \to \frac{\pi}{4}} \; \left[\tan 2x \log \left(\tan x\right)\right] & \left(\infty - 0 \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left[\dfrac{\log \left(\tan x\right)}{\cot 2x}\right] & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}}\; \dfrac{\dfrac{d}{dx}\left[\log \left(\tan x\right)\right]}{\dfrac{d}{dx}\left(\cot 2x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sec^2 x / \tan x}{-2 \text{ cosec}^2 \; 2x} & \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\dfrac{1}{\cos^2 x} \div \dfrac{\sin x}{\cos x}}{\dfrac{-2}{\sin^2 2x}} & \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin^2 2x}{-2 \sin x \cos x} & \\ & & \\ & = - \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin^2 2x}{\sin 2x} & \left[\text{Note: }\sin 2x = 2 \sin x \cos x\right] \\ & & \\ & = - \lim\limits_{x \to \frac{\pi}{4}} \; \sin 2x & \\ & & \\ & = - \sin \left(2 \times \dfrac{\pi}{4}\right) & \\ & & \\ & = - \sin \dfrac{\pi}{2} = -1 \end{aligned}$

$\therefore$ $\;$ $\lim\limits_{x \to \frac{\pi}{4}} \; \left\{\log \left[f\left(x\right)\right] \right\} = -1$

i.e. $\log \left\{\lim\limits_{x \to \frac{\pi}{4}} \left[f\left(x\right)\right] \right\} = -1$

i.e. $\lim\limits_{x \to \frac{\pi}{4}} \left(\tan x\right)^{\tan 2x} = e^{-1} = \dfrac{1}{e}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{\sin^{-1}x- \tan^{-1}x}{x^3}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\sin^{-1}x - \tan^{-1}x}{x^3} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin^{-1}x - \tan^{-1}x\right)}{\dfrac{d}{dx}\left(x^3\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{1+x^2}}{3x^2} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{1+x^2}\right)}{\dfrac{d}{dx}\left(3x^2\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{-\dfrac{1}{2}\times \left(1-x^2\right)^{-3/2}\times \left(-2x\right)+\dfrac{2x}{\left(1+x^2\right)^2}}{6x} & \\ & & \\ & = \dfrac{1}{6} \; \lim\limits_{x \to 0} \; \left\{\dfrac{1}{\left(1-x^2\right)^{3/2}} + \dfrac{2}{\left(1+x^2\right)^2} \right\} & \\ & & \\ & = \dfrac{1}{6} \times 3 = \dfrac{1}{2} \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{\cos x -1}{\cos 2x -1}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\cos x -1}{\cos 2x -1} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\cos x -1\right)}{\dfrac{d}{dx}\left(\cos 2x -1\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{-\sin x}{-2\sin 2x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin x\right)}{\dfrac{d}{dx}\left(2 \sin 2x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\cos x}{4 \cos 2x} & \\ & & \\ & = \dfrac{1}{4} & \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{4^x - 9^x}{x \left(4^x + 9^x\right)}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{4^x - 9^x}{x \left(4^x + 9^x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(4^x - 9^x\right)}{\dfrac{d}{dx}\left[x\left(4^x + 9^x\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{4^x \ln 4 - 9^x \ln 9}{4^x + 9^x + x \left(4^x \ln 4 + 9^x \ln 9\right)} \\ & & \\ & = \dfrac{\ln 4 - \ln 9}{1+1+0} \\ & & \\ & = \dfrac{1}{2} \ln \left(\dfrac{4}{9}\right) \\ & & \\ & = \dfrac{1}{2} \ln \left(\dfrac{2}{3}\right)^2 \\ & & \\ & = \dfrac{1}{2} \times 2 \ln \left(\dfrac{2}{3}\right) = \ln \left(\dfrac{2}{3}\right) \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to a} \; \dfrac{\log \left(x-a\right)}{\log \left(e^x - e^a\right)}$

$\begin{aligned} & \lim\limits_{x \to a} \; \dfrac{\log \left(x-a\right)}{\log \left(e^x - e^a\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx} \left[\log \left(x-a\right)\right]}{\dfrac{d}{dx} \left[\log \left(e^x - e^a\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{1}{x-a}}{\dfrac{e^x}{e^x - e^a}} \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{e^x - e^a}{e^x \left(x-a\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx}\left(e^x - e^a\right)}{\dfrac{d}{dx}\left[e^x \left(x-a\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{e^x}{e^x \left(x-a\right)+e^x} \\ & & \\ & = \dfrac{e^a}{e^a} = 1 \end{aligned}$

Limits Indeterminate Form

If $\lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} = -1$, find the value of a.

$\begin{aligned} & \lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx}\left(a^x - x^a\right)}{\dfrac{d}{dx}\left(x^x - a^a\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{a^x \ln a - ax^{a-1}}{x^x \left(1+ \ln x\right)} & \\ & & \\ & = \dfrac{a^a \ln a -a \times a^{a-1}}{a^a \left(1+ \ln a\right)} & \\ & & \\ & = \dfrac{a^a \ln a -a^a}{a^a \left(1+ \ln a\right)} & \\ & & \\ & = \dfrac{\ln a -1}{\ln a + 1} & \end{aligned}$

$\therefore$ $\;$ $\lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} = -1$ $\implies$ $\dfrac{\ln a - 1}{\ln a + 1}=-1$

i.e. $\ln a -1 = -\ln a - 1$

i.e. $2 \ln a = 0$

i.e. $a = e^0 = 1$