Question

Use l'Hôpital's Rule to evaluate the following limits. $$\underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}x}{\tan (\pi x/2)}$$

Short answer

Based on the provided step-by-step solution, evaluate the given limit: $$\underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}x}{\tan (\pi x/2)}$$ The limit of the given expression is: $$L=\underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}x}{\tan (\pi x/2)}=0$$

Step-by-step solution

Check if the limit is in the form $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$

As x approaches 1 from the left side, let's find out what happens to the numerator and the denominator of the given fraction. Numerator: $$\underset{x\to 1^{-}}{lim}{\tanh }^{-1}x$$ At x=1, ${\tanh }^{-1}x={\tanh }^{-1}(1)=0$ Denominator: $$\underset{x\to 1^{-}}{lim}\tan (\pi x/2)$$ At x=1, $\tan (\pi x/2)=\tan (\pi /2)=\mathrm{\infty }$ Since the limit is not in the form $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$, l'Hôpital's Rule cannot be applied in this case. To evaluate this limit, we'll need to use a different technique.

Apply Squeeze Theorem

Consider the inequality: $$-1\le x\le 1$$ Take hyperbolic arctangent of all three terms in the inequality: $${\tanh }^{-1}(-1)\le {\tanh }^{-1}(x)\le {\tanh }^{-1}(1)$$ We know that: $${\tanh }^{-1}(-1)=-\frac{\pi }{4}\text{and}{\tanh }^{-1}(1)=\frac{\pi }{4}$$ Divide the inequality by $\tan (\pi x/2)$: $$\frac{{\tanh }^{-1}(-1)}{\tan (\pi x/2)}\le \frac{{\tanh }^{-1}(x)}{\tan (\pi x/2)}\le \frac{{\tanh }^{-1}(1)}{\tan (\pi x/2)}$$ Now, take the limit of the inequality as x approaches 1 from the left side: $$\underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}(-1)}{\tan (\pi x/2)}\le \underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}(x)}{\tan (\pi x/2)}\le \underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}(1)}{\tan (\pi x/2)}$$ The limit of the middle term is the limit we're looking for: $$L=\underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}(x)}{\tan (\pi x/2)}$$ Let's find the limits of the outer terms: Left outer term: Since ${\tanh }^{-1}(-1)=-\frac{\pi }{4}$, we have: $$\underset{x\to 1^{-}}{lim}\frac{-\frac{\pi }{4}}{\tan (\pi x/2)}=\underset{x\to 1^{-}}{lim}-\frac{\pi }{4\cdot \tan (\pi x/2)}=0$$ Right outer term: Since ${\tanh }^{-1}(1)=\frac{\pi }{4}$, we have: $$\underset{x\to 1^{-}}{lim}\frac{\frac{\pi }{4}}{\tan (\pi x/2)}=\underset{x\to 1^{-}}{lim}\frac{\pi }{4\cdot \tan (\pi x/2)}=0$$ Now, using the Squeeze Theorem, we have: $$0\le L\le 0$$ Thus, the limit of the given expression is: $$L=\underset{x\to 1^{-}}{lim}\frac{{\tanh }^{-1}x}{\tan (\pi x/2)}=0$$

Key concepts

Squeeze Theorem

The Squeeze Theorem is a fundamental concept in calculus used to evaluate limits that are difficult to determine directly. It is particularly useful when dealing with functions that are "squeezed" between two other functions whose limits are known. If you have a function, say $f(x)$, and it satisfies the condition:

• $g(x)\le f(x)\le h(x)$

for all $x$ in some interval, except possibly at the point where $x$ approaches a, and

• $\underset{x\to a}{lim}g(x)=\underset{x\to a}{lim}h(x)=L$

then the Squeeze Theorem tells us that:

• $\underset{x\to a}{lim}f(x)=L$

This theorem allows us to determine the limit of $f(x)$ by "squeezing" it between two simpler functions with known limits. In our exercise, the Squeeze Theorem was utilized to find the limit of a function involving the hyperbolic arctangent and tangent functions by examining the function as $x$ approaches 1 from the left. By showing that both bounding functions approached 0, we concluded that the target function also approached 0.

Limit Evaluation

Limit evaluation is a core part of calculus, which involves finding the value that a function approaches as the input approaches a certain point. It's an essential concept to understand continuity and the behavior of functions at specific points.

There are several methods to evaluate limits:

• Direct substitution, where you plug the value of $x$ into the function. This works when the function is continuous at that point.
• Factoring, useful for resolving expressions that initially give an indeterminate form, like $\frac{0}{0}$.
• Rationalization, commonly used when dealing with limits involving square roots.
• L'Hôpital's Rule, applied to indeterminate forms like $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$.
• The Squeeze Theorem, as explained earlier, helpful when a function is "squeezed" between two other functions.

In the original exercise, L'Hôpital's Rule was not applicable, but the Squeeze Theorem provided the means to accurately evaluate the limit, demonstrating how diverse techniques can be paired depending on the situation.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but based on hyperbolas rather than circles. Just as sine and cosine originate from the unit circle, hyperbolic sine ($\sinh$) and hyperbolic cosine ($\cosh$) derive from the hyperbola.

Important hyperbolic functions include:

• $\sinh (x)=\frac{e^x-e^{-x}}{2}$
• $\cosh (x)=\frac{e^x+e^{-x}}{2}$
• Other functions like $\tanh (x)$, $\text{coth}(x)$, $\text{sech}(x)$, and $\text{csch}(x)$ are derived from these.

Particularly, the inverse hyperbolic tangent function, ${\tanh }^{-1}(x)$, plays a critical role in understanding limits as it approaches certain points. In the given exercise, ${\tanh }^{-1}(x)$ was essential in applying the Squeeze Theorem to accurately assess the limit. Understanding these functions expands the toolkit available for analyzing complex mathematical problems.