Calculating Limit of Function – A function with e in the power of a function to zero – Exercise 6319

Exercise

Evaluate the following limit:

$\lim _ { x \rightarrow 0} {(e^x+3x)}^{\frac{1}{x}}$

Final Answer

Show final answer

Solution

First, we try to plug in $x = 0$ and get

${(e^0+3\cdot 0)}^{\frac{1}{0}}$

We got the phrase $1^0$ (=tending to 1 in the power of tending to 0). This is an indeterminate form, therefore we have to get out of this situation.

$\lim _ { x \rightarrow 0} {(e^x+3x)}^{\frac{1}{x}}=$

Using Logarithm Rules we get

$=\lim _ { x \rightarrow 0} e^{\ln {(e^x+3x)}^{\frac{1}{x}}}=$

$=\lim _ { x \rightarrow 0} e^{\frac{1}{x}\ln (e^x+3x)}=$

We simplify the phrase and get

$=\lim _ { x \rightarrow 0} e^{\frac{\ln (e^x+3x)}{x}}=$

We enter the limit inside

$= e^{\lim _ { x \rightarrow 0} \frac{\ln (e^x+3x)}{x}}=$

Note: This can be done because an exponential function is a continuous function.

We plug in 0 again and get

$=e^{\frac{\ln (e^0+3\cdot 0)}{0}}=$

$=e^{\frac{\ln 1}{0}}=$

$=e^{\frac{0}{0}}=$

We got the phrase $\frac{"0"}{"0"}$(=tending to zero divides tending to zero). This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get

$= e^{\lim _ { x \rightarrow 0} \frac{\frac{1}{e^x+3x}\cdot (e^x+3)}{1}}=$

We simplify the phrase and get

$= e^{\lim _ { x \rightarrow 0} \frac{e^x+3}{e^x+3x}}=$

We plug in zero again and get

$= e^{ \frac{e^0+3}{e^0+3\cdot 0}}=$

$= e^{ \frac{1+3}{1+0}}=$

$= e^4$

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