L'Hôpital's Rule.

Theorem (L'Hôpital's Rule).

Let functions $f(x)$ and $g(x)$:

a) be differentiable in the vicinity of point $a$, except possibly at point $a$, where $g'(x)\neq 0$ in this neighborhood;

b) simultaneously, functions $f(x)$ and $g(x)$ are either both infinitesimal or both infinite as $x\rightarrow a$;

c) there exists a finite $\lim\limits_{x\rightarrow a}\frac{f'(x)}{g'(x)}.$

Then $\lim\limits_{x\rightarrow a}\frac{f(x)}{g(x)}$ exists, and the equality $\lim\limits_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim\limits_{x\rightarrow a}\frac{f'(x)}{g'(x)}$ holds.

If functions $f(x)$ and $g(x)$ are differentiable at point $a$,

$g(a)=f(a)=0,$ $ g'(a)\neq 0$, then $\lim\limits_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{f'(a)}{g'(a)}.$

Examples:

1. $\lim\limits_{x\rightarrow 1}\frac{x^5-1}{2x^3-x-1}$

We have an indeterminate form $\frac{0}{0}.$ Applying L'Hôpital's Rule, we obtain:

$\lim\limits_{x\rightarrow 1}\frac{x^5-1}{2x^3-x-1}=\lim\limits_{x\rightarrow 1}\frac{5x^4}{6x^2-1}=1.$

2. $\lim\limits_{x\rightarrow 0}\frac{x-\arctan x}{x^3}$

We have an indeterminate form $\frac{0}{0}.$ Applying L'Hôpital's Rule, we obtain:

$$\lim\limits_{x\rightarrow 0}\frac{x-arctg x}{x^3}=\lim\limits_{x\rightarrow 0}\frac{1-\frac{1}{1+x^2}}{3x^2}=\lim\limits_{x\rightarrow 0}\frac{x^2}{3x^2(1+x^2)=\frac{1}{3}.}$$

3. $\lim\limits_{x\rightarrow 1}\frac{\ln x}{\sqrt{x}}$

We have an indeterminate form $\frac{\infty}{\infty}.$ Applying L'Hôpital's Rule, we obtain:

$$\lim\limits_{x\rightarrow +\infty}\frac{\ln x}{\sqrt{x}}=\lim\limits_{x\rightarrow +\infty}\frac{1/x}{1/(2\sqrt{x})}=\lim\limits_{x\rightarrow +\infty}\frac{2}{\sqrt{x}}=0.$$

4. $\lim\limits_{x\rightarrow 0}\frac{\sin x-x\cos x}{\sin^3 x}.$

We have an indeterminate form $\frac{0}{0}.$ Recognizing that $\sin x\sim x$ as $x\rightarrow 0$, by L'Hôpital's Rule, we find

$\lim\limits_{x\rightarrow 0}\frac{\sin x-x\cos x}{\sin^3 x}=\lim\limits_{x\rightarrow 0}\frac{\sin x-x\cos x}{x^3}=\lim\limits_{x\rightarrow 0}\frac{\cos x-\cos x+x\sin x}{3x^2}=$ $\frac{1}{3}\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=\frac{1}{3}.$

5. $\lim\limits_{x\rightarrow 1}\frac{x^10-10x+9}{x^5-5x+4}.$

We have an indeterminate form $\frac{0}{0}.$ Applying L'Hôpital's Rule, we get:

$\lim\limits_{x\rightarrow 1}\frac{x^{10}-10x+9}{x^5-5x+4}=\lim\limits_{x\rightarrow 1}\frac{10x^9-10}{5x^4-5}.$

Using L'Hôpital's Rule again, we find

$\lim\limits_{x\rightarrow 1}\frac{10x^9-10}{5x^4-5}=2\lim\limits_{x\rightarrow 1}\frac{x^9-1}{x^4-1}=2\lim\limits_{x\rightarrow 1}\frac{9x^8}{4x^3}=\frac{9}{2}.$

6. $\lim\limits_{x\rightarrow+\infty}\frac{x^{\alpha}}{e^{\beta x}},$ где $\alpha>0,$ $\beta>0.$

Let $k=[\alpha]+1;$ then $\alpha-k<0.$

Applying L'Hôpital's Rule $k$ times, we obtain $\lim\limits_{x\rightarrow+\infty}\frac{x^{\alpha}}{e^{\beta x}}=\lim\limits_{x\rightarrow+\infty}\frac{\alpha x^{\alpha-1}}{\beta e^{\beta x}}=...=\lim\limits_{x\rightarrow+\infty}\frac{\alpha(\alpha-1)...(\alpha-k+1)x^{\alpha-k}}{\beta^k e^{\beta x}}=0.$

7. $\lim\limits_{x\rightarrow+\infty}\frac{\ln^{\alpha}x}{x^{\beta}},$ where $\alpha>0,$ $\beta>0.$

Let $\ln x =t;$ then $x=e^t$ and $\lim\limits_{x\rightarrow+\infty}\frac{\ln^{\alpha}x}{x^{\beta}}=\lim\limits_{t\rightarrow+\infty}\frac{t^{\alpha}}{e^{\beta t}}=0$ (example 6).

We have an indeterminate form $\frac{0}{0}.$ Applying L'Hôpital's Rule, we get:

8. $\lim\limits_{x\rightarrow +0}x\ln x$

By transforming the $0\cdot\infty$ indeterminate form to $\frac{\infty}{\infty}$ form and applying L'Hôpital's Rule, we have

$$\lim\limits_{x\rightarrow +0}x\ln x=\lim\limits_{x\rightarrow +0}\frac{\ln x}{1/x}=\lim\limits_{x\rightarrow +0}\frac{1/x}{-1/x^2}=\lim\limits_{x\rightarrow +0}(-x)=0.$$

9. $\lim\limits_{x\rightarrow 1}\frac{1}{x^{50}}e^{-1/x^2}.$

We have an indeterminate form of $\frac{0}{0}.$ Assuming $1/x^2=t,$ we get

$$\lim\limits_{x\rightarrow 1}\frac{1}{x^{50}}e^{-1/x^2}=\lim\limits_{x\rightarrow 1}\frac{t^{25}}{e^t}=0.$$

10. $\lim\limits_{x\rightarrow 0}\left(\frac{1}{x^2}-ctg^2 x\right).$

Transforming the indeterminate form $\infty-\infty$ to the form $\frac{0}{0}$ and using the asymptotic formula $\sin x \sim x$ as $x$ approaches $0$, we get

$$\lim\limits_{x\rightarrow 0}\left(\frac{1}{x^2}-ctg^2 x\right)=\lim\limits_{x\rightarrow 0}\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}=$$ $$=\lim\limits_{x\rightarrow 0}\frac{(\sin x+x\cos x)(\sin x-x\cos x)}{x^2\sin^2 x}=$$ $$=\lim\limits_{x\rightarrow 0}\frac{\sin x+x\cos x}{x}\cdot\lim\limits_{x\rightarrow 0}\frac{\sin x-x\cos x}{x^3}.$$

Since

$$\lim\limits_{x\rightarrow 0}\frac{\sin x+x\cos x}{x}=\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}+\lim\limits_{x\rightarrow 0}\cos x=2,$$

and $\lim\limits_{x\rightarrow 0}\frac{\sin x-x\cos x}{x^3}=\frac{1}{3}$ (see Example 4), the required limit is $2/3$.

Tags: L'Hôpital's Rule, calculus, differential, higher-order differentials, l'hopital's rule, mathematical analysis