Question

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

lim_{x\to0}\frac{x2^{x}}{2^{x}-1}

resources
read it watch it

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, \(\lim_{x
ightarrow0}\frac{x2^{x}}{2^{x}-1}=\lim_{x
ightarrow0}\frac{2^{x}+x2^{x}\ln2}{2^{x}\ln2}\).

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:

\(\frac{1}{\ln2}\)