L'Hopital's Theorem

L'Hopital's theorem is a powerful tool used in calculus to evaluate limits that are of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It was first introduced by French mathematician Guillaume de l'Hopital in 1696 in his book "Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes".

Statement of the Theorem

If $f(x)$ and $g(x)$ are two functions such that:

1. $f(x)$ and $g(x)$ are differentiable in a neighborhood of $a$, except possibly at $a$ itself.
2. $g'(x)\neq 0$ in a neighborhood of $a$, except possibly at $a$ itself.
3. $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=0$ or $\lim_{x\to a}f(x)=\pm\infty$ and $\lim_{x\to a}g(x)=\pm\infty$.

Then,

$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$

provided the limit of the right-hand side exists or is infinite.

Intuition behind the Theorem

The idea behind L'Hopital's theorem is that if the limit of $\frac{f(x)}{g(x)}$ is indeterminate ($\frac{0}{0}$ or $\frac{\infty}{\infty}$), then the limit of $\frac{f'(x)}{g'(x)}$ will often provide a more manageable expression to evaluate the limit. This is because the derivative of a function can often simplify or cancel out terms that were causing the original expression to be indeterminate.

Examples

Let's look at some examples of how L'Hopital's theorem can be used to evaluate limits.

Example 1

Evaluate $\lim_{x\to 0}\frac{\sin x}{x}$.

This limit is of the form $\frac{0}{0}$, so we can apply L'Hopital's theorem. Taking the derivative of both the numerator and denominator, we get:

$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\cos x}{1}=1$

Example 2

Evaluate $\lim_{x\to\infty}\frac{x}{\ln x}$.

This limit is of the form $\frac{\infty}{-\infty}$ (as $\ln x$ approaches $\infty$ as $x$ approaches $\infty$), so we can apply L'Hopital's theorem. Taking the derivative of both the numerator and denominator, we get:

$\lim_{x\to\infty}\frac{x}{\ln x}=\lim_{x\to\infty}\frac{1}{1/x}= \lim_{x\to\infty}x=\infty$

Example 3

Evaluate $\lim_{x\to 0^+}x\ln x$.

This limit is of the form $0\cdot-\infty$, so we can apply L'Hopital's theorem. Taking the derivative of both the numerator and denominator, we get:

$\lim_{x\to 0^+}x\ln x=\lim_{x\to 0^+}\frac{\ln x}{1/x}=\lim_{x\to 0^+}\frac{1/x}{-1/x^2}=\lim_{x\to 0^+}-x=0$

Conclusion

L'Hopital's theorem is a powerful tool for evaluating limits that are initially indeterminate. It is important to note that the theorem only applies in certain circumstances, and it is always a good idea to check the conditions before applying it. With practice, L'Hopital's theorem can become a valuable tool in the calculus toolkit.

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