Exercise
Evaluate the following limit:
\lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}}
Final Answer
Solution
First, we try to plug in x = \infty and get
\frac{8^\infty}{{(4+\frac{2}{\infty})}^{\frac{3\infty}{2}}}
We got the phrase \frac{\infty}{\infty} (=infinity divides by infinity). This is an indeterminate form, therefore we have to get out of this situation.
We will simplify the phrase using Powers and Roots Rules:
\lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}}=
=\lim _ { x \rightarrow \infty} \frac{2^{3x}}{{(4(1+\frac{2}{4x}))}^{\frac{3x}{2}}}=
=\lim _ { x \rightarrow \infty} \frac{2^{3x}}{{(2^2)}^{\frac{3x}{2}}{(1+\frac{1}{2x})}^{\frac{3x}{2}}}=
=\lim _ { x \rightarrow \infty} \frac{2^{3x}}{2^{3x}{(1+\frac{1}{2x})}^{\frac{3x}{2}}}=
=\lim _ { x \rightarrow \infty} \frac{1}{{(1+\frac{1}{2x})}^{\frac{3x}{2}}}=
=\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{-\frac{3x}{2}}=
We will plug in infinity again and get
={(1+\frac{1}{2\infty})}^{-\frac{3\infty}{2}}=1^{-\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. At the base, we have an expression of the form:
1+\frac{1}{2x}
And the following holds:
\lim _ { x \rightarrow \infty} \frac{1}{2x}= 0
As required.
Therefore, we will multiply the power by the inverted expression and get
=\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x\cdot\frac{1}{2x}\cdot (-\frac{3x}{2})}=
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+\frac{1}{2x})}^{2x}=e
All we have left to do is to calculate the limit on the expression that remains in the power:
\lim _ { x \rightarrow \infty}\frac{1}{2x}\cdot (-\frac{3x}{2})=
=\lim _ { x \rightarrow \infty}\frac{-3x}{4x}=
=-\frac{3}{4}
Therefore, in total we get
=\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x\cdot\frac{1}{2x}\cdot (-\frac{3x}{2})}=
=e^{-\frac{3}{4}}
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