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Chapter 4: Problem 65

Evaluate the following limits or explain why they do not exist. Check your results by graphing. $$\lim _{x \rightarrow 0^{+}}(\tan x)^{x}$$

Short Answer

Expert verified
Question: Evaluate the limit or explain why it does not exist for the function \((\tan x)^{x}\) as \(x\) approaches \(0^{+}\). Answer: The limit as \(x\) approaches \(0^{+}\) for the function \((\tan x)^{x}\) is \(1\).

Step by step solution

01

Rewrite the expression

The given limit is: $$\lim_{x \rightarrow 0^{+}}(\tan x)^{x}$$ We can rewrite this expression by taking the natural logarithm to help us simplify: $$\lim_{x \rightarrow 0^{+}} e ^{\ln(\tan x)^x}$$
02

Use logarithm properties

Use properties of logarithms to rewrite the limit: $$\lim_{x \rightarrow 0^{+}} e^{\displaystyle x \ln(\tan x)}$$
03

Use L'H么pital's Rule

Since we have \(\lim_{x \rightarrow 0^{+}} \frac{x \ln (\tan x)}{1}\), a limit of the form \(0 \cdot \infty\), we can rewrite it as a fraction so we can apply L'H么pital's Rule: $$\lim_{x \rightarrow 0^{+}} e^{\frac{\ln(\tan x)}{\frac{1}{x}}}$$ Now with the indeterminate form \(\frac{0}{0}\) after substitution, we can apply L'H么pital's Rule by explaining the limit as: $$\lim_{x \rightarrow 0^{+}} e^{\frac{\frac{1}{\tan x} \cdot \sec ^2 x}{-\frac{1}{x ^2}}}$$
04

Simplify the expression and find the limit

Simplify the expression and substitute \(x=0\) to find the actual limit: $$\lim_{x \rightarrow 0^{+}} e^{\frac{x^2 \cdot \sec^2 x}{\tan x}} = e^{\frac{0^2 \cdot 1}{0}} = e^0$$ Since any number raised to the power of \(0\) is equal to \(1\): $$e^0 = 1$$
05

Conclusion

Thus, based on the steps and simplification, the limit for the given function is: $$\lim _{x \rightarrow 0^{+}}(\tan x)^{x} = 1$$

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