Question

question 4 (1 point)

\\( \lim _{x \
ightarrow 0^{+}} 10 x \ln (x)= \\)

\\( -\infty \\)

0

\\( \pi \\)

1

\\( \infty \\)

Explanation:

Step1: Rewrite the limit

We have \(\lim_{x
ightarrow0^{+}}10x\ln(x)\). Let \(t = \frac{1}{x}\), then as \(x
ightarrow0^{+}\), \(t
ightarrow+\infty\). And \(x\ln(x)=\frac{\ln(\frac{1}{t})}{t}=\frac{-\ln(t)}{t}\). So the original limit becomes \(10\lim_{t
ightarrow+\infty}\frac{-\ln(t)}{t}\).

Step2: Apply L - H rule

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

Step3: Evaluate the limit

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

Answer:

\(0\)