Question

d) $limlimits_{x\toinfty}(e^{2x}+3)^{3/x}$

Explanation:

Step1: Take the natural logarithm

Let \( y=(e^{2x}+3)^{\frac{3}{x}} \), then \( \ln y=\frac{3}{x}\ln(e^{2x}+3) \).

Step2: Use L'Hopital's Rule

As \( x
ightarrow\infty \), \( \frac{\ln(e^{2x}+3)}{\frac{x}{3}} \) is in the \( \frac{\infty}{\infty} \) form.
Differentiate the numerator and denominator:
The derivative of \( \ln(e^{2x}+3) \) is \( \frac{2e^{2x}}{e^{2x}+3} \), and the derivative of \( \frac{x}{3} \) is \( \frac{1}{3} \).
So \( \lim_{x
ightarrow\infty}\ln y = 3\lim_{x
ightarrow\infty}\frac{2e^{2x}}{e^{2x}+3} \).
Divide numerator and denominator by \( e^{2x} \): \( 3\lim_{x
ightarrow\infty}\frac{2}{1 + 3e^{-2x}} \).
Since \( \lim_{x
ightarrow\infty}e^{-2x}=0 \), then \( 3\times\frac{2}{1+0}=6 \).

Step3: Find the original limit

Since \( \lim_{x
ightarrow\infty}\ln y = 6 \), then \( \lim_{x
ightarrow\infty}y = e^{6} \).

Answer:

\( e^{6} \)