Question

f. $limlimits_{x \to 0}\frac{x - sin x}{x^{3}+x^{2}}$
g. $limlimits_{x \to 0}(1 + 2x)^{1/x}$
h. $limlimits_{x \to 0^{+}}x^{2}ln(x)$
i. $limlimits_{x \to infty}(x^{2}+1)^{1/ln x}$
j. $limlimits_{x \to 0^{+}}x^{x}$

Explanation:

Step1: Use L'Hopital's Rule for F

For $\lim_{x
ightarrow0}\frac{x - \sin x}{x^{3}+x^{2}}$, it is in $\frac{0}{0}$ form.
Differentiate numerator and denominator:
$\lim_{x
ightarrow0}\frac{1-\cos x}{3x^{2}+2x}$ (still $\frac{0}{0}$)
Differentiate again:
$\lim_{x
ightarrow0}\frac{\sin x}{6x + 2}=\frac{0}{2}=0$

Step2: Rewrite G

Let $y=(1 + 2x)^{1/x}$, then $\ln y=\frac{\ln(1 + 2x)}{x}$
$\lim_{x
ightarrow0}\ln y=\lim_{x
ightarrow0}\frac{\ln(1 + 2x)}{x}$ ( $\frac{0}{0}$ form)
Using L'Hopital's Rule: $\lim_{x
ightarrow0}\frac{\frac{2}{1+2x}}{1}=2$
So $\lim_{x
ightarrow0}(1 + 2x)^{1/x}=e^{2}$

Step3: Rewrite H

$\lim_{x
ightarrow0^{+}}x^{2}\ln x=\lim_{x
ightarrow0^{+}}\frac{\ln x}{x^{-2}}$ ( $\frac{-\infty}{\infty}$ form)
Using L'Hopital's Rule: $\lim_{x
ightarrow0^{+}}\frac{\frac{1}{x}}{-2x^{-3}}=\lim_{x
ightarrow0^{+}}\frac{-x^{2}}{2}=0$

Step4: Rewrite I

Let $y=(x^{2}+1)^{1/\ln x}$, then $\ln y=\frac{\ln(x^{2}+1)}{\ln x}$
$\lim_{x
ightarrow\infty}\ln y=\lim_{x
ightarrow\infty}\frac{\frac{2x}{x^{2}+1}}{\frac{1}{x}}=\lim_{x
ightarrow\infty}\frac{2x^{2}}{x^{2}+1}=2$
So $\lim_{x
ightarrow\infty}(x^{2}+1)^{1/\ln x}=e^{2}$

Step5: Rewrite J

Let $y = x^{x}$, then $\ln y=x\ln x=\frac{\ln x}{x^{-1}}$ ( $\frac{-\infty}{\infty}$ form as $x
ightarrow0^{+}$)
Using L'Hopital's Rule: $\lim_{x
ightarrow0^{+}}\ln y=\lim_{x
ightarrow0^{+}}\frac{\frac{1}{x}}{-x^{-2}}=\lim_{x
ightarrow0^{+}}(-x)=0$
So $\lim_{x
ightarrow0^{+}}x^{x}=e^{0}=1$

Answer:

F. $0$; G. $e^{2}$; H. $0$; I. $e^{2}$; J. $1$