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Chapter 6: Problem 87

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

Short Answer

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

Step by step solution

01

Write the limit using natural logarithm

Since we want to use l'H么pital'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 \(L\).
02

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'H么pital's Rule, we need to convert this product into a fraction. We can use the following identity: $$xy = \frac{x}{\frac{1}{y}}$$
03

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'H么pital's Rule.
04

Apply l'H么pital's Rule

Applying l'H么pital'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: $$\frac{\mathrm{d}}{\mathrm{d} x}\ln(\tanh x) = \frac{1}{\tanh x}\cdot \frac{\mathrm{d}}{\mathrm{d} x}\tanh x = \frac{1}{\tanh x}\cdot\operatorname{sech}^2 x$$
05

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: $$\ln(L) = 0$$
06

Solve for L

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

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