Problem 84 - Calculus Early Transcendentals

Question

Use l'Hôpital's Rule to evaluate the following limits. $$\lim _{x \rightarrow \infty} \frac{1-\operatorname{coth} x}{1-\tanh x}$$

Step-by-Step Solution

Answer

Answer: The value of the limit is 1.

Step 1: Differentiate the numerator and denominator

Differentiate the expressions \(1-\operatorname{coth} x\) and \(1-\tanh x\) separately with respect to \(x\). The derivatives are: $$\frac{d}{dx}(1-\coth x) = -\csch^2 x$$ $$\frac{d}{dx}(1-\tanh x) = -\operatorname{sech}^2 x$$

Step 2: Apply l'Hôpital's Rule

Since we have the indeterminate form \(\frac{0}{0}\), we can apply l'Hôpital's Rule. This involves taking the limit of the ratio of the derivatives of the functions. So, the limit becomes: $$\lim _{x \rightarrow \infty} \frac{-\csch^2 x}{-\operatorname{sech}^2 x}$$

Step 3: Calculate the limit

Now, we have to find the limit of the new expression as \(x\) approaches infinity. We can simplify the expression by cancelling out the negative signs: $$\lim_{x \rightarrow \infty} \frac{\csch^2 x}{\operatorname{sech}^2 x}$$ We can find the equivalent expressions for \(\csch x\) and \(\operatorname{sech} x\) using \(\sinh x\) and \(\cosh x\), respectively: $$\lim_{x \rightarrow \infty} \frac{(\frac{1}{\sinh x})^2}{(\frac{1}{\cosh x})^2} = \lim_{x \rightarrow \infty} \frac{\cosh^2 x}{\sinh^2 x}$$ Now, recall the identity for hyperbolic functions, \(\cosh^2 x - \sinh^2 x = 1\). We can rewrite the limit expression as: $$\lim_{x \rightarrow \infty} \frac{\sinh^2 x + 1}{\sinh^2 x}$$ Now, we can divide both the numerator and denominator by \(\sinh^2 x\), so the expression becomes: $$\lim_{x \rightarrow \infty} \frac{1 + \frac{1}{\sinh^2 x}}{1}$$ As \(x\) approaches infinity, \(\frac{1}{\sinh^2 x}\) approaches zero. Therefore, the limit is: $$\lim_{x \rightarrow \infty} \frac{1 + 0}{1} = \boxed{1}$$ So, the value of the limit using l'Hôpital's Rule is 1.