# Calculating Limit of Function – A function with e in the power of a function with e to infinity – Exercise 6329

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow \infty} {(1+2e^{-x})}^{e^x+x}$$

$$\lim _ { x \rightarrow \infty} {(1+2e^{-x})}^{e^x+x}=e^2$$

### Solution

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

$${(1+2e^{-\infty})}^{e^\infty+\infty}$$

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

We will use the known limit also called Euler’s Limit.

$$\lim _ { x \rightarrow \infty} {(1+2e^{-x})}^{e^x+x}=$$

At the base, we have an expression of the form:

$$1+2e^{-x}$$

And the following holds:

$$\lim _ { x \rightarrow \infty} 2e^{-x}= 0$$

As required.

Therefore, we will multiply the power by the inverted expression and get

$$=\lim _ { x \rightarrow \infty} {(1+2e^{-x})}^{\frac{1}{2e^{-x}}\cdot 2e^{-x}\cdot (e^x+x)}=$$

Note: When adding an expression by multiplication, its inverted expression must also be multiplied so that the original expression does not change.

Now, according to Euler’s Limit, we get

$$\lim _ { x \rightarrow \infty} {(1+2e^{-x})}^{\frac{1}{2e^{-x}}}=e$$

All we have left to do is to calculate the limit on the expression that remains in the power:

$$\lim _ { x \rightarrow \infty} 2e^{-x}\cdot (e^x+x)=$$

$$=2\lim _ { x \rightarrow \infty} \frac{e^x+x}{e^x}=$$

We plug in infinity again and get

$$=2\cdot\frac{e^\infty+\infty}{e^\infty}=\frac{\infty}{\infty}$$

We got the phrase $$\frac{\infty}{\infty}$$. This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get

$$=2\lim _ { x \rightarrow \infty} \frac{e^x+1}{e^x}=$$

$$=2\lim _ { x \rightarrow \infty} 1+\frac{1}{e^x}=$$

We plug in infinity again and this time we get

$$=2\cdot(1+\frac{1}{e^\infty})=$$

$$=2\cdot(1+\frac{1}{\infty})=$$

$$=2\cdot(1+0)=$$

$$=2$$

Therefore, in total we get

$$\lim _ { x \rightarrow \infty} {(1+2e^{-x})}^{\frac{1}{2e^{-x}}\cdot 2e^{-x}\cdot (e^x+x)}=$$

$$=e^2$$

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!

Share with Friends