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:

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

Final Answer

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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

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

$=e^2$

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