Exercise
Evaluate the following limit:
\lim _ { x \rightarrow \infty} {(1+2e^{-x})}^{e^x+x}
Final Answer
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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