find the following limit.\n\n\\( \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { 5 } \\ln x \\)\n\n\\( \\lim _ {…

find the following limit.\n\n\\( \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { 5 } \\ln x \\)\n\n\\( \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { 5 } \\ln x = \\) 0 (simplify your answer.)

find the following limit.\n\n\\( \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { 5 } \\ln x \\)\n\n\\( \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { 5 } \\ln x = \\) 0 (simplify your answer.)

Answer

Explanation:

Step1: Rewrite the limit

We have (\lim_{x\rightarrow0^{+}}x^{5}\ln x). This is in the (0\times(-\infty)) indeterminate form. We can rewrite it as (\lim_{x\rightarrow0^{+}}\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\rightarrow0^{+}}\frac{\ln x}{x^{-5}}=\lim_{x\rightarrow0^{+}}\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\rightarrow0^{+}}-\frac{x^{5}}{5}). Substitute (x = 0) into (-\frac{x^{5}}{5}). Since (x\rightarrow0^{+}), when (x = 0), (-\frac{x^{5}}{5}=0).

Answer:

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