Question

1. find the limit.

\\( \lim _ { x \
ightarrow 0 } \frac { x \left( 2 ^ { x } \
ight) } { 2 ^ { x } - 1 } \\)

Explanation:

Step1: Check the form

When \(x = 0\), \(\frac{x(2^{x})}{2^{x}-1}=\frac{0\times(2^{0})}{2^{0}-1}=\frac{0}{0}\), which is an indeterminate form. So we can use L - H rule.

Step2: Apply L - H rule

Differentiate the numerator and denominator.
The derivative of \(y = x\cdot2^{x}\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) (\(u = x\), \(u^\prime=1\), \(v = 2^{x}\), \(v^\prime=2^{x}\ln2\)) is \(2^{x}+x\cdot2^{x}\ln2\).
The derivative of \(y = 2^{x}-1\) is \(2^{x}\ln2\).
So \(\lim_{x
ightarrow0}\frac{x(2^{x})}{2^{x}-1}=\lim_{x
ightarrow0}\frac{2^{x}+x\cdot2^{x}\ln2}{2^{x}\ln2}\).

Step3: Substitute \(x = 0\)

Substitute \(x = 0\) into \(\frac{2^{x}+x\cdot2^{x}\ln2}{2^{x}\ln2}\).
When \(x = 0\), \(\frac{2^{0}+0\times2^{0}\ln2}{2^{0}\ln2}=\frac{1 + 0}{1\times\ln2}=\frac{1}{\ln2}\).

Answer:

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