Question

use lhopitals rule to find the limit.

$$\lim_{x \to 0} \frac{x 11^{x}}{11^{x}-1}$$

$$\lim_{x \to 0} \frac{x 11^{x}}{11^{x}-1}=\square$$ (type an exact answer.)

Explanation:

Step1: Check form

When \(x = 0\), \(\frac{x11^{x}}{11^{x}-1}=\frac{0\times11^{0}}{11^{0}-1}=\frac{0}{0}\), so L'Hôpital's Rule applies.

Step2: Differentiate numerator and denominator

Differentiate \(y = x11^{x}\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) (\(u = x\), \(v = 11^{x}\), \(u^\prime=1\), \(v^\prime = 11^{x}\ln11\)), we get \(y^\prime=11^{x}+x11^{x}\ln11\).
Differentiate \(y = 11^{x}-1\), we get \(y^\prime=11^{x}\ln11\).
So \(\lim_{x
ightarrow0}\frac{x11^{x}}{11^{x}-1}=\lim_{x
ightarrow0}\frac{11^{x}+x11^{x}\ln11}{11^{x}\ln11}\).

Step3: Substitute \(x = 0\)

Substitute \(x = 0\) into \(\frac{11^{x}+x11^{x}\ln11}{11^{x}\ln11}\), we have \(\frac{11^{0}+0\times11^{0}\ln11}{11^{0}\ln11}=\frac{1 + 0}{\ln11}=\frac{1}{\ln11}\).

Answer:

\(\frac{1}{\ln11}\)