Question

9 multiple choice 3 points evaluate: \\( \lim _{x \
ightarrow 0^{+}} \sin (x) \ln (x) \\) 1 \\( \infty \\) \\( -\infty \\) 0 the limit does not exist 10 multiple choice 3 points evaluate: \\( \lim _{x \
ightarrow \infty}\left(e^{x}+5 x\
ight)^{\frac{1}{4 x}} \\) 0 1 \\( e \\) \\( e^{1 / 4} \\) \\( \infty \\)

Explanation:

Step1: Use the product - limit rule and known limits

We know that \(\lim_{x
ightarrow0^{+}}\sin(x)=0\) and \(\lim_{x
ightarrow0^{+}}\ln(x)=-\infty\). But we can rewrite \(\sin(x)\ln(x)\) as \(\frac{\ln(x)}{\frac{1}{\sin(x)}}\). Then, by L'Hopital's rule (since we have \(-\infty/\infty\) form), \(\lim_{x
ightarrow0^{+}}\frac{\ln(x)}{\csc(x)}\).
Differentiate the numerator and denominator: \(\frac{d}{dx}\ln(x)=\frac{1}{x}\) and \(\frac{d}{dx}\csc(x)=-\csc(x)\cot(x)\). So the limit becomes \(\lim_{x
ightarrow0^{+}}\frac{\frac{1}{x}}{-\csc(x)\cot(x)}=\lim_{x
ightarrow0^{+}}\frac{-\sin^{2}(x)}{x\cos(x)}\).

Step2: Use small - angle approximations

As \(x
ightarrow0^{+}\), \(\sin(x)\approx x\). Then \(\lim_{x
ightarrow0^{+}}\frac{-\sin^{2}(x)}{x\cos(x)}=\lim_{x
ightarrow0^{+}}\frac{-x^{2}}{x\cos(x)}=\lim_{x
ightarrow0^{+}}\frac{-x}{\cos(x)}\).

Step3: Evaluate the limit

Substitute \(x = 0\) into \(\frac{-x}{\cos(x)}\). Since \(\cos(0)=1\), \(\lim_{x
ightarrow0^{+}}\frac{-x}{\cos(x)} = 0\).

for problem 10:

Step1: Take the natural logarithm

Let \(y=(e^{x}+5x)^{\frac{1}{4x}}\). Then \(\ln(y)=\frac{\ln(e^{x}+5x)}{4x}\).

Step2: Use L'Hopital's rule

As \(x
ightarrow\infty\), we have \(\frac{\infty}{\infty}\) form. Differentiate the numerator and denominator: \(\frac{d}{dx}\ln(e^{x}+5x)=\frac{e^{x}+5}{e^{x}+5x}\) and \(\frac{d}{dx}(4x) = 4\).
\(\lim_{x
ightarrow\infty}\frac{e^{x}+5}{4(e^{x}+5x)}\). Divide numerator and denominator by \(e^{x}\): \(\lim_{x
ightarrow\infty}\frac{1 + 5e^{-x}}{4(1+5xe^{-x})}\).

Step3: Evaluate the limit

Since \(\lim_{x
ightarrow\infty}e^{-x}=0\) and \(\lim_{x
ightarrow\infty}xe^{-x}=0\) (using L'Hopital's rule on \(\lim_{x
ightarrow\infty}\frac{x}{e^{x}}\) with \(\frac{\infty}{\infty}\) form, \(\lim_{x
ightarrow\infty}\frac{1}{e^{x}} = 0\)), \(\lim_{x
ightarrow\infty}\frac{1 + 5e^{-x}}{4(1+5xe^{-x})}=\frac{1}{4}\).

Step4: Find the original limit

Since \(\lim_{x
ightarrow\infty}\ln(y)=\frac{1}{4}\), then \(\lim_{x
ightarrow\infty}y = e^{\frac{1}{4}}\) (because \(y = e^{\ln(y)}\)).

Answer:

\(0\)