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 $g^{\prime }(x)\ne 0$ in this neighborhood;

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

c) there exists a finite $\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}.$

Then $\underset{x\to a}{lim}\frac{f(x)}{g(x)}$ exists, and the equality $\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ holds.

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

$g(a)=f(a)=0,$ $g^{\prime }(a)\ne 0$, then $\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\frac{f^{\prime }(a)}{g^{\prime }(a)}.$

Examples:

1. $\underset{x\to 1}{lim}\frac{x^5-1}{2x^3-x-1}$

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

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

2. $\underset{x\to 0}{lim}\frac{x-\arctan x}{x^3}$

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

$$\underset{x\to 0}{lim}\frac{x-arctgx}{x^3}=\underset{x\to 0}{lim}\frac{1-\frac{1}{1+x^2}}{3x^2}=\underset{x\to 0}{lim}\frac{x^2}{3x^2(1+x^2)=\frac{1}{3}.}$$

3. $\underset{x\to 1}{lim}\frac{\ln x}{\sqrt{x}}$

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

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{\ln x}{\sqrt{x}}=\underset{x\to +\mathrm{\infty }}{lim}\frac{1/x}{1/(2\sqrt{x})}=\underset{x\to +\mathrm{\infty }}{lim}\frac{2}{\sqrt{x}}=0.$$

4. $\underset{x\to 0}{lim}\frac{\sin x-x\cos x}{{\sin }^{3}x}.$

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

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

5. $\underset{x\to 1}{lim}\frac{x^{1}0-10x+9}{x^5-5x+4}.$

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

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

Using L'Hôpital's Rule again, we find

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

6. $\underset{x\to +\mathrm{\infty }}{lim}\frac{x^{\alpha }}{e^{\beta x}},$ где $\alpha >0,$ $\beta >0.$

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

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

7. $\underset{x\to +\mathrm{\infty }}{lim}\frac{{\ln }^{\alpha }x}{x^{\beta }},$ where $\alpha >0,$ $\beta >0.$

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

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

8. $\underset{x\to +0}{lim}x\ln x$

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

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

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

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

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

10. $\underset{x\to 0}{lim}(\frac{1}{x^2}-ct{g}^{2}x).$

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

$$\underset{x\to 0}{lim}(\frac{1}{x^2}-ct{g}^{2}x)=\underset{x\to 0}{lim}\frac{{\sin }^{2}x-x^2{\cos }^{2}x}{x^2{\sin }^{2}x}=$$ $$=\underset{x\to 0}{lim}\frac{(\sin x+x\cos x)(\sin x-x\cos x)}{x^2{\sin }^{2}x}=$$ $$=\underset{x\to 0}{lim}\frac{\sin x+x\cos x}{x}\cdot \underset{x\to 0}{lim}\frac{\sin x-x\cos x}{x^3}.$$

Since

$$\underset{x\to 0}{lim}\frac{\sin x+x\cos x}{x}=\underset{x\to 0}{lim}\frac{\sin x}{x}+\underset{x\to 0}{lim}\cos x=2,$$

and $\underset{x\to 0}{lim}\frac{\sin x-x\cos x}{x^3}=\frac{1}{3}$ (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