f. $$ \\lim _ { x \\rightarrow 0 } \\frac { x - \\sin x } { x ^ { 3 } + x ^ { 2 } } $$\ng. $$ \\lim _ { x…

f. $$ \\lim _ { x \\rightarrow 0 } \\frac { x - \\sin x } { x ^ { 3 } + x ^ { 2 } } $$\ng. $$ \\lim _ { x \\rightarrow 0 } ( 1 + 2 x ) ^ { 1 / x } $$\nh. $$ \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { 2 } \\ln ( x ) $$\ni. $$ \\lim _ { x \\rightarrow \\infty } \\left( x ^ { 2 } + 1 \\right) ^ { 1 / \\ln x } $$\nj. $$ \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { x } $$

f. $$ \\lim _ { x \\rightarrow 0 } \\frac { x - \\sin x } { x ^ { 3 } + x ^ { 2 } } $$\ng. $$ \\lim _ { x \\rightarrow 0 } ( 1 + 2 x ) ^ { 1 / x } $$\nh. $$ \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { 2 } \\ln ( x ) $$\ni. $$ \\lim _ { x \\rightarrow \\infty } \\left( x ^ { 2 } + 1 \\right) ^ { 1 / \\ln x } $$\nj. $$ \\lim _ { x \\rightarrow 0 ^ { + } } x ^ { x } $$

Answer

Explanation:

Step1: Apply L'Hospital's Rule for F

For $\lim_{x\rightarrow0}\frac{x - \sin x}{x^{3}+x^{2}}$, it is in $\frac{0}{0}$ form. Differentiate numerator and denominator: $$\lim_{x\rightarrow0}\frac{1-\cos x}{3x^{2}+2x}$$ Still $\frac{0}{0}$ form. Differentiate again: $$\lim_{x\rightarrow0}\frac{\sin x}{6x + 2}=\frac{0}{2}=0$$

Step2: Use the formula for G

For $\lim_{x\rightarrow0}(1 + 2x)^{1/x}$, let $y=(1 + 2x)^{1/x}$. Then $\ln y=\frac{\ln(1 + 2x)}{x}$. Apply L'Hospital's Rule (since $\frac{0}{0}$ form): $$\lim_{x\rightarrow0}\frac{\frac{2}{1+2x}}{1}=2$$ So $\lim_{x\rightarrow0}y = e^{2}$

Step3: Rewrite for H

For $\lim_{x\rightarrow0^{+}}x^{2}\ln x$, rewrite as $\lim_{x\rightarrow0^{+}}\frac{\ln x}{x^{-2}}$. It is in $\frac{-\infty}{\infty}$ form. Apply L'Hospital's Rule: $$\lim_{x\rightarrow0^{+}}\frac{\frac{1}{x}}{-2x^{-3}}=\lim_{x\rightarrow0^{+}}\frac{-x^{2}}{2}=0$$

Step4: Take natural log for I

Let $y=(x^{2}+1)^{1/\ln x}$. Then $\ln y=\frac{\ln(x^{2}+1)}{\ln x}$. Apply L'Hospital's Rule (as $x\rightarrow\infty$, $\frac{\infty}{\infty}$ form): $$\lim_{x\rightarrow\infty}\frac{\frac{2x}{x^{2}+1}}{\frac{1}{x}}=\lim_{x\rightarrow\infty}\frac{2x^{2}}{x^{2}+1}=2$$ So $\lim_{x\rightarrow\infty}y = e^{2}$

Step5: Rewrite for J

Let $y = x^{x}$. Then $\ln y=x\ln x=\frac{\ln x}{x^{-1}}$. As $x\rightarrow0^{+}$, $\frac{-\infty}{\infty}$ form. Apply L'Hospital's Rule: $$\lim_{x\rightarrow0^{+}}\frac{\frac{1}{x}}{-x^{-2}}=\lim_{x\rightarrow0^{+}}(-x)=0$$ So $\lim_{x\rightarrow0^{+}}y = e^{0}=1$

Answer:

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