L Hospital’s Rule is used to find the limiting values of some functions which take indeterminate forms. Let’s dive into more detailed explanation of L Hospital’s Rule.
Let us consider two functions $ƒ(x)=x^3-8$ and $g(x)=x^2-4$. If we take the limit of these two functions when $x\to 2$, we get \[\lim_{x\to 2}ƒ(x)=0 \text{ and } \lim_{x\to 2}g(x)=0\] \[\therefore \lim_{x\to 2}\frac{ƒ(x)}{g(x)}=\frac{0}{0}\]
This form $\frac{0}{0}$ is known as indeterminate form. We have other indeterminate forms $\frac{\infty}{\infty}$, $\infty-\infty$ as well. But here, we consider only the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$. L Hospital’s rule is used to find the limiting values of these functions which take the form $\frac{0}{0}$ and $\frac{\infty}{\infty}$ when $x\to a$.
If $ƒ(x)$ and $g(x)$ and their derivatives $ƒ'(x)$ and $g'(x)$ are continuous at $x=a$ [Continuity], and if $ƒ(a)=g(a)=0$, then \[\lim_{x\to a}\frac{ƒ(x)}{g(x)}=\frac{\lim_{x\to a}ƒ'(x)}{\lim_{x\to a}g'(x)}=\frac{ƒ'(a)}{g'(a)}\] provided that $g'(a)≠0$. If $ƒ'(a)$ and $g'(a)$ are both zero then above theorem can further be used.
Thus \[\lim_{x\to a}\frac{ƒ'(x)}{g'(x)}=\lim_{x\to a}\frac{ƒ^{\prime\prime}(x)}{g^{\prime\prime}(x)}=\frac{ƒ^{\prime\prime}(a)}{g^{\prime\prime}(a)}\] provided that $ƒ^{\prime\prime}(x)$ and $g^{\prime\prime}(x)$ both are continuos and $g^{\prime\prime}(a)≠0$. The above theorem can further be used if $ƒ^{\prime\prime}(a)=0$ and $g^{\prime\prime}(a)=0$.
If $\lim_{x\to a}ƒ(x)=\infty$ and $\lim_{x\to a}g(x)=\infty$ then $\lim_{x\to a}\frac{ƒ(x)}{g(x)}=\frac{\infty}{\infty}$. This form $\frac{\infty}{\infty}$ can be solved by expressing $\frac{ƒ(x)}{g(x)}$ into $\frac{\frac{1}{g(x)}}{\frac{1}{ƒ(x)}}$ so that each of the numerator and denominator will tend to zero as $x\to a$. Hence, the form $\frac{\infty}{\infty}$ changes into the form $\frac{0}{0}$ after which L Hospital’s rule can be applied to evaluate the limit.
Example \[\lim_{x\to a}\frac{x^n-a^n}{x-a} \text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to a} \frac{nx^{n-1}}{1}=na^{n-1}\] \[\therefore \lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}\]
In the same way, the above illustrated problem can be solved as \[\lim_{x\to 2}\frac{x^3-8}{x^2-4}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to 2}\frac{3x^2}{2x}\] \[=\frac{3×2^2}{2×2}=3\]
Example of the form $\frac{\infty}{\infty}$ \[\lim_{x\to\infty}\frac{x^3}{e^x} \text{ (form }\frac{\infty}{\infty}\text{)}\] \[=\lim_{x\to\infty}\frac{3x^2}{e^x}\text{ (form }\frac{\infty}{\infty}\text{)}\] \[=\lim_{x\to\infty}\frac{6x}{e^x} \text{ (form }\frac{\infty}{\infty}\text{)}\] \[=\lim_{x\to\infty}\frac{6}{e^x}=0\]
Evaluate $\lim_{x\to 0}\frac{e^x+e^{-x}-2\cos x}{\sin^2x}$
\[\lim_{x\to 0}\frac{e^x+e^{-x}-2\cos x}{\sin^2x}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to 0}\frac{e^x-e^{-x}+2\sin x}{2\sin x\cos x}\] \[=\lim_{x\to 0} \frac{e^x-e^{-x}+2\sin x}{\sin 2x}\] \[=\lim_{x\to 0}\frac{e^x+e^{-x}+2\cos x}{2\cos 2x}\] \[=\frac{e^0+e^0+2\cos 0}{2\cos 0}=\frac{1+1+2}{2}=2\]
Evaluate $\lim_{x\to 1}\frac{1-2x+x^2}{1+\log x-x}$
\[\lim_{x\to 1}\frac{1-2x+x^2}{1+\log x-x}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to 1}\frac{-2+2x}{-1+\frac{1}{x}}\] \[\lim_{x\to 1}\frac{2}{-\frac{1}{x^2}}\] \[=\lim_{x\to 1}-2x^2=-2×1^2=-2\]
Evaluate $\lim_{x\to 0}\frac{\tan x-x}{x-\sin x}$
\[\lim_{x\to 0}\frac{\tan x-x}{x-\sin x}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to 0}\frac{\sec^x-1}{1-\cos x}\] \[\lim_{x\to 0}\frac{\tan^2x}{1-\cos x}\] \[=\lim_{x\to 0}\frac{2\tan x\sec^2x}{\sin x}\] \[=\lim_{x\to 0}2\sec^3x=2\sec^30=2\]
Evaluate $\lim_{x\to 0}\frac{x-\sin x\cos x}{x^3}$
\[\lim_{x\to a}\frac{x-\sin x\cos x}{x^3}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to 0}\frac{2x-2\sin x\cos x}{2x^3}\] \[=\lim_{x\to 0}\frac{2x-\sin 2x}{2x^3}\] \[=\lim_{x\to 0}\frac{2-2\cos 2x}{6x^2}\] \[=\frac{2}{3}\lim_{x\to 0}\frac{1-\cos 2x}{2x^2}\] \[=\frac{2}{3}\lim_{x\to 0}\frac{2\sin^2x}{2x^2}\] \[=\frac{2}{3}\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2=\frac{2}{3}\]
Evaluate $\lim_{x\to 0}\frac{(e^x-1)\tan x}{x^2}$
\[\lim_{x\to 0}\frac{(e^x-1)\tan x}{x^2}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to 0}\frac{e^x-1}{x}\lim_{x\to 0}\frac{\tan x}{x}=1\]
Evaluate $\lim_{x\to \infty}\frac{3x^2-5}{2x^3+4x+3}$
\[\lim_{x\to\infty}\frac{3x^2-5}{2x^3+4x+3}\text{ (form }\frac{\infty}{\infty}\text{)}\] \[=\lim_{x\to\infty}\frac{6x}{6x^2+4}\] \[=\lim_{x\to\infty}\frac{6x}{6x^2+4}\] \[=\lim_{x\to\infty}\frac{6}{12x}=0\]
Evaluate $\lim_{x\to\frac{π}{2}}\frac{\sec 7x}{\sec 5x}$
\[\lim_{x\to\frac{π}{2}}\frac{\sec 7x}{\sec 5x}\text{ (form }\frac{\infty}{\infty}\text{)}\] \[=\lim_{x\to\frac{π}{2}}\frac{\cos 5x}{\cos 7x}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to\frac{π}{2}}\frac{-5\sin 5x}{-7\sin 7x}\] \[=\frac{5}{7}\frac{\sin\frac{5π}{2}}{\sin\frac{7π}{2}}\] \[=\frac{5}{7}×\frac{1}{-1}=\frac{-5}{7}\]
Evaluate $\lim_{x\to 0}\frac{\log x}{\log\cot x}$
\[\lim_{x\to 0}\frac{\log x}{\log\cot x}\text{ (form }\frac{\infty}{\infty}\text{)}\] \[=\lim_{x\to 0}\frac{\frac{1}{x}}{\frac{1}{\cot x}×(-\operatorname{cosec}^2x)}\] \[=-\lim_{x\to 0}\frac{\cot x}{x\operatorname{cosec}^2x}\] \[=-\lim_{x\to 0}\frac{\cos x\sin x}{x}\] \[=-\lim_{x\to 0}\frac{2\cos x\sin x}{2x}\] \[=-\lim_{x\to 0}\frac{\sin 2x}{2x}\] \[=-\lim_{x\to 0}\frac{2\cos 2x}{2}\] \[=-\lim_{x\to 0}\cos 2x=-\cos 0=-1\]
Evaluate $\lim_{x\to\frac{π}{2}}\frac{\tan 5x}{\tan x}$
\[\lim_{x\to\frac{π}{2}}\frac{\tan 5x}{\tan x}\text{ (form }\frac{\infty}{\infty}\text{)}\] \[=\lim_{x\to\frac{π}{2}}\frac{\cot x}{\cot 5x}\text{ (form }\frac{0}{0}\text{)}\] \[=\lim_{x\to\frac{π}{2}}\frac{-\operatorname{cosec}^2x}{-5\operatorname{cosec}^25x}\] \[=\frac{1}{5}\frac{\operatorname{cosec}^2\frac{π}{2}}{\operatorname{cosec}^2\frac{5π}{2}}=\frac{1}{5}\]