Problem 87 - Calculus Early Transcendentals

Question

Use l'Hôpital's Rule to evaluate the following limits. $$\lim _{x \rightarrow 0^{+}}(\tanh x)^{x}$$

Step-by-Step Solution

Answer

Question: Evaluate the limit: $$\lim _{x \rightarrow 0^{+}}(\tanh x)^{x}$$ Answer: The given limit is equal to 1.

Step 1: Write the limit using natural logarithm

Since we want to use l'Hôpital's Rule, we should rewrite the given limit using the natural logarithm. Let's denote the given limit as \(L\). We can write $$L = \lim _{x \rightarrow 0^{+}}(\tanh x)^{x}$$ Taking the natural logarithm of both sides: $$\ln(L) = \lim _{x \rightarrow 0^{+}}x \ln(\tanh x)$$ The goal now is to find the limit on the right-hand side, and then exponentiate the result to find \(L\).

Step 2: Identify indeterminate form

We now have the limit in the form of a product. As \(x\) approaches \(0^+\), the first term \(x\) approaches \(0\), and the second term \(\ln(\tanh x)\) approaches \(-\infty\) since \(\tanh(0) = 0\). So we have an indeterminate form \(0 \times (-\infty)\). To apply l'Hôpital's Rule, we need to convert this product into a fraction. We can use the following identity: $$xy = \frac{x}{\frac{1}{y}}$$

Step 3: Rewrite as a fraction

Rewriting the limit using the identity mentioned in Step 2, we have: $$\ln(L) = \lim _{x \rightarrow 0^{+}}\frac{x}{\frac{1}{\ln(\tanh x)}}$$ As \(x \to 0^+\), we have the indeterminate form \(0/0\), so we can apply l'Hôpital's Rule.

Step 4: Apply l'Hôpital's Rule

Applying l'Hôpital's Rule, we differentiate the numerator and the denominator with respect to \(x\). We get: $$\ln(L) = \lim _{x \rightarrow 0^{+}}\frac{1}{-\frac{1}{\ln(\tanh x)^2} \cdot \frac{\mathrm{d} (\ln(\tanh x))}{\mathrm{d} x}}$$ Now we need to find the derivative of \(\ln(\tanh x)\) with respect to \(x\). Using the chain rule, we get: $$\frac{\mathrm{d}}{\mathrm{d} x}\ln(\tanh x) = \frac{1}{\tanh x}\cdot \frac{\mathrm{d}}{\mathrm{d} x}\tanh x = \frac{1}{\tanh x}\cdot\operatorname{sech}^2 x$$

Step 5: Plug in derivative and simplify

Replacing the derivative in the limit expression, we have: $$\ln(L) = \lim _{x \rightarrow 0^{+}}\frac{1}{-\frac{1}{\ln(\tanh x)^2} \cdot \frac{1}{\tanh x}\cdot\operatorname{sech}^2 x}$$ Simplifying further, we get: $$\ln(L) = \lim _{x \rightarrow 0^{+}} -\ln(\tanh x)^2 \cdot \tanh x \cdot\operatorname{sech}^2 x$$ As \(x \to 0^+\), \(\ln(\tanh x)^2 \to 0\), \(\tanh x \to 0\), and \(\operatorname{sech}^2 x \to 1\). Thus, we have: $$\ln(L) = 0$$

Step 6: Solve for L

Now we can find \(L\) by exponentiating both sides: $$L = e^{\ln(L)} = e^0 = 1$$ So the limit is: $$\lim _{x \rightarrow 0^{+}}(\tanh x)^{x} = 1$$