find the limit. use lhospitals rule where appropriate. if there is a more elementary consider using…

find the limit. use lhospitals rule where appropriate. if there is a more elementary consider using it.\n\nlim_{x\\to0}\\frac{x2^{x}}{2^{x}-1}\n\nresources\nread it watch it

find the limit. use lhospitals rule where appropriate. if there is a more elementary consider using it.\n\nlim_{x\\to0}\\frac{x2^{x}}{2^{x}-1}\n\nresources\nread it watch it

Answer

Explanation:

Step1: Check the form of the limit

When (x = 0), we have (\frac{0\times2^{0}}{2^{0}-1}=\frac{0}{0}) (indeterminate form). So, we can apply L'Hospital's Rule.

Step2: Differentiate numerator and denominator

Differentiate (y = x2^{x}) using the product rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = x), (u^\prime=1), (v = 2^{x}), (v^\prime=2^{x}\ln2). So, ((x2^{x})^\prime=2^{x}+x2^{x}\ln2). Differentiate (y = 2^{x}-1), ((2^{x}-1)^\prime=2^{x}\ln2). Now, (\lim_{x\rightarrow0}\frac{x2^{x}}{2^{x}-1}=\lim_{x\rightarrow0}\frac{2^{x}+x2^{x}\ln2}{2^{x}\ln2}).

Step3: Substitute (x = 0)

Substitute (x = 0) into (\frac{2^{x}+x2^{x}\ln2}{2^{x}\ln2}). When (x = 0), (2^{0}=1), and we get (\frac{1 + 0\times1\times\ln2}{1\times\ln2}=\frac{1}{\ln2}).

Answer:

(\frac{1}{\ln2})