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


Evaluate the following limit:

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

Final Answer

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


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


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:


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


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





Therefore, in total we get

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


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