Question

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

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

1. -/1 points

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

\\( \lim _ { x \
ightarrow 0 ^ { + } } \frac { \arctan ( 8 x ) } { \ln ( x ) } \\)

Explanation:

Step1: Check the form of the limit

When \(x
ightarrow0^{+}\), \(\arctan(8x)
ightarrow0\) and \(\ln(x)
ightarrow-\infty\). The limit \(\lim_{x
ightarrow0^{+}}\frac{\arctan(8x)}{\ln(x)}\) is in the form \(\frac{0}{-\infty}\).

Step2: Apply the property of limits

We know that if \(\lim_{x
ightarrow a}f(x) = 0\) and \(\lim_{x
ightarrow a}|g(x)|=\infty\), then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)} = 0\). Here \(f(x)=\arctan(8x)\), \(g(x)=\ln(x)\) and \(a = 0^{+}\).

Answer:

\(0\)