use lhopitals rule to find the limit.\n\n$$\\lim_{x \\to 0} \\frac{x 11^{x}}{11^{x}-1}$$\n\n$$\\lim_{x \\to…

use lhopitals rule to find the limit.\n\n$$\\lim_{x \\to 0} \\frac{x 11^{x}}{11^{x}-1}$$\n\n$$\\lim_{x \\to 0} \\frac{x 11^{x}}{11^{x}-1}=\\square$$ (type an exact answer.)

use lhopitals rule to find the limit.\n\n$$\\lim_{x \\to 0} \\frac{x 11^{x}}{11^{x}-1}$$\n\n$$\\lim_{x \\to 0} \\frac{x 11^{x}}{11^{x}-1}=\\square$$ (type an exact answer.)

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\rightarrow0}\frac{x11^{x}}{11^{x}-1}=\lim_{x\rightarrow0}\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})