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

### Exercise

Evaluate the following limit:

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

$$\lim _ { x \rightarrow 0} {(e^x+3x)}^{\frac{1}{x}}=e^4$$

### 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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