Question

1. $lim_{x

ightarrow+infty}\frac{e^{-x}}{ln(1 + 4e^{-x})}$

Explanation:

Step1: Let \(t = e^{-x}\)

As \(x
ightarrow+\infty\), then \(t
ightarrow0^{+}\). The limit becomes \(\lim_{t
ightarrow0^{+}}\frac{t}{\ln(1 + 4t)}\).

Step2: Use L - H rule

Since \(\lim_{t
ightarrow0^{+}}t=0\) and \(\lim_{t
ightarrow0^{+}}\ln(1 + 4t)=0\), and both are differentiable. The derivative of \(y = t\) is \(y^\prime=1\), and the derivative of \(y=\ln(1 + 4t)\) is \(y^\prime=\frac{4}{1 + 4t}\) by the chain - rule.

Step3: Calculate the new limit

\(\lim_{t
ightarrow0^{+}}\frac{t}{\ln(1 + 4t)}=\lim_{t
ightarrow0^{+}}\frac{1}{\frac{4}{1 + 4t}}=\lim_{t
ightarrow0^{+}}\frac{1 + 4t}{4}\).

Step4: Evaluate the limit

Substitute \(t = 0\) into \(\frac{1+4t}{4}\), we get \(\frac{1+4\times0}{4}=\frac{1}{4}\).

Answer:

\(\frac{1}{4}\)