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Chapter 4: Problem 63

Evaluate the following limits or explain why they do not exist. Check your results by graphing. $$\lim _{x \rightarrow 0}\left(e^{a x}+x\right)^{1 / x}, \text { for a constant } a$$

Short Answer

Expert verified
Question: Evaluate the limit of the function $$\lim _{x \rightarrow 0}\left(e^{a x}+x\right)^{1 / x}$$, where 'a' is a constant. Answer: The limit of the function as x approaches 0 is $$e^{a + \frac{1+a}{2}}$$.

Step by step solution

01

Simplify the expression using limit properties

First, rewrite the expression using properties of exponents: $$\lim _{x \rightarrow 0}\left(e^{a x}+x\right)^{1 / x} = \lim _{x \rightarrow 0} e^{\frac{a x + \ln(x + e^{a x})}{x}}$$
02

Apply L'Hopital's Rule

Now, we need to find the limit of the exponent, i.e., find the limit as x approaches 0 for the following expression: $$\frac{a x + \ln(x + e^{a x})}{x}$$ As it is in the form of 0/0, we can apply L'Hopital's rule. Differentiate both the numerator and the denominator with respect to x: $$\frac{a + \frac{1 + ae^{ax}}{1 + e^{ax}}}{1}$$ Now, we can evaluate the limit as x approaches 0: $$a + \frac{1 + ae^{a * 0}}{1 + e^{a * 0}}= a + \frac{1+a}{1+1}= a+\frac{1+a}{2}$$
03

Substitute back the limit

Now that we have found the limit of the exponent, we can substitute it back into the expression: $$\lim_{x \rightarrow 0}\left(e^{a x}+x\right)^{1 / x} = e^{a + \frac{1+a}{2}}$$ So, the limit is: $$\lim _{x \rightarrow 0}\left(e^{a x}+x\right)^{1 / x} = e^{a + \frac{1+a}{2}}$$
04

Graph the function and check the limit

To check the result using graphing, use a graphing calculator or an online graphing tool like Desmos or GeoGebra. Graph the function $$y=\left(e^{a x}+x\right)^{1 / x}$$ and observe that as x approaches 0, the expression approaches the value, $$e^{a + \frac{1+a}{2}}$$, confirming our result.

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