Exercise
Evaluate the following limit:
\lim _ { x \rightarrow \infty} {(\frac{2x-3}{2x+4})}^{\frac{x-1}{4}}
Final Answer
Solution
First, we try to plug in x = \infty
{(\frac{2\infty-3}{2\infty+4})}^{\frac{\infty-1}{4}}
In the base 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 simplify the expression in order to use the known limit also called Euler’s Limit.
\lim _ { x \rightarrow \infty} {(\frac{2x-3}{2x+4})}^{\frac{x-1}{4}}=
=\lim _ { x \rightarrow \infty} {(\frac{2x+4-7}{2x+4})}^{\frac{x-1}{4}}=
=\lim _ { x \rightarrow \infty} {(1+\frac{-7}{2x+4})}^{\frac{x-1}{4}}=
At the base, we have an expression of the form:
1+\frac{-7}{2x+4}
And the following holds:
\lim _ { x \rightarrow \infty} \frac{-7}{2x+4}= 0
As required.
Therefore, we will multiply the power by the inverted expression and get
=\lim _ { x \rightarrow \infty} {(1+\frac{-7}{2x+4})}^{\frac{2x+4}{-7}\cdot\frac{-7}{2x+4}\cdot\frac{x-1}{4}}=
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{-7}{2x+4})}^{\frac{2x+4}{-7}}=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{-7}{2x+4}\cdot\frac{x-1}{4}=
=\lim _ { x \rightarrow \infty}\frac{-7(x-1)}{4(2x+4)}=
=\lim _ { x \rightarrow \infty}\frac{-7x+7}{8x+16}=
We received a quotient of polynomials tending to infinity. In such a situation, we divide the numerator and denominator by the expression with highest power, without its coefficient. We will get
=\lim _ { x \rightarrow \infty}\frac{\frac{-7x+7}{x}}{\frac{8x+16}{x}}=
=\lim _ { x \rightarrow \infty}\frac{-7+\frac{7}{x}}{8+\frac{16}{x}}=
We plug in infinity and get
=\frac{-7+\frac{7}{\infty}}{8+\frac{16}{\infty}}=
=\frac{-7+0}{8+0}=
=\frac{-7}{8}
Therefore, in total we get
=\lim _ { x \rightarrow \infty} {(1+\frac{-7}{2x+4})}^{\frac{2x+4}{-7}\cdot\frac{-7}{2x+4}\cdot\frac{x-1}{4}}=
=e^{-\frac{7}{8}}
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