question 4 (1 point)\n\n\\( \\lim _{x \\rightarrow 0^{+}} 10 x \\ln (x)= \\)\n\n\\( -\\infty \\)\n\n0\n\n\\(…

Explanation:

Step1: Rewrite the limit

We have (\lim_{x\rightarrow0^{+}}10x\ln(x)). Let (t = \frac{1}{x}), then as (x\rightarrow0^{+}), (t\rightarrow+\infty). And (x\ln(x)=\frac{\ln(\frac{1}{t})}{t}=\frac{-\ln(t)}{t}). So the original limit becomes (10\lim_{t\rightarrow+\infty}\frac{-\ln(t)}{t}).

Step2: Apply L - H rule

Since (\lim_{t\rightarrow+\infty}\frac{-\ln(t)}{t}) is in the (\frac{\infty}{\infty}) form. By L - H rule ((\lim_{t\rightarrow a}\frac{f(t)}{g(t)}=\lim_{t\rightarrow a}\frac{f^{\prime}(t)}{g^{\prime}(t)}) when (\lim_{t\rightarrow a}f(t)=\lim_{t\rightarrow a}g(t)=\pm\infty)), where (f(t)=-\ln(t)), (f^{\prime}(t)=-\frac{1}{t}) and (g(t) = t), (g^{\prime}(t)=1). Then (\lim_{t\rightarrow+\infty}\frac{-\ln(t)}{t}=\lim_{t\rightarrow+\infty}\frac{-\frac{1}{t}}{1}).

Step3: Evaluate the limit

(\lim_{t\rightarrow+\infty}\frac{-\frac{1}{t}}{1}=0) (because (\lim_{t\rightarrow+\infty}\frac{1}{t} = 0)). So (10\lim_{t\rightarrow+\infty}\frac{-\ln(t)}{t}=0).

Answer:

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