Question

find the following limit.

\\( \lim _ { x \
ightarrow 0 ^ { + } } x ^ { 5 } \ln x \\)

\\( \lim _ { x \
ightarrow 0 ^ { + } } x ^ { 5 } \ln x = \\) 0 (simplify your answer.)

Explanation:

Step1: Rewrite the limit

We have \(\lim_{x
ightarrow0^{+}}x^{5}\ln x\). This is in the \(0\times(-\infty)\) indeterminate form. We can rewrite it as \(\lim_{x
ightarrow0^{+}}\frac{\ln x}{x^{- 5}}\) (now in the \(\frac{-\infty}{\infty}\) indeterminate form, so we can apply L'Hopital's Rule).

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of \(\ln x\) is \(\frac{1}{x}\), and the derivative of \(x^{-5}\) is \(-5x^{-6}\). So, \(\lim_{x
ightarrow0^{+}}\frac{\ln x}{x^{-5}}=\lim_{x
ightarrow0^{+}}\frac{\frac{1}{x}}{- 5x^{-6}}\).
Simplify the expression: \(\frac{\frac{1}{x}}{-5x^{-6}}=\frac{1}{x}\times\frac{x^{6}}{-5}=-\frac{x^{5}}{5}\).

Step3: Evaluate the new limit

Now, find \(\lim_{x
ightarrow0^{+}}-\frac{x^{5}}{5}\). Substitute \(x = 0\) into \(-\frac{x^{5}}{5}\). Since \(x
ightarrow0^{+}\), when \(x = 0\), \(-\frac{x^{5}}{5}=0\).

Answer:

\(0\)