f. $\\lim_{x\\to0}\\frac{x - \\sin x}{x^{3}+x^{2}}$ \ng. $\\lim_{x\\to0}(1 + 2x)^{1/x}$ \nh…

f. $\\lim_{x\\to0}\\frac{x - \\sin x}{x^{3}+x^{2}}$ \ng. $\\lim_{x\\to0}(1 + 2x)^{1/x}$ \nh. $\\lim_{x\\to0^{+}}x^{2}\\ln(x)$ \ni. $\\lim_{x\\to\\infty}(x^{2}+1)^{1/\\ln x}$ \nj. $\\lim_{x\\to0^{+}}x^{x}$

f. $\\lim_{x\\to0}\\frac{x - \\sin x}{x^{3}+x^{2}}$ \ng. $\\lim_{x\\to0}(1 + 2x)^{1/x}$ \nh. $\\lim_{x\\to0^{+}}x^{2}\\ln(x)$ \ni. $\\lim_{x\\to\\infty}(x^{2}+1)^{1/\\ln x}$ \nj. $\\lim_{x\\to0^{+}}x^{x}$

Answer

Explanation:

Step1: Apply L'Hopital's Rule

Since $\lim_{x\rightarrow0}\frac{x - \sin x}{x^{3}+x^{2}}$ is in the $\frac{0}{0}$ form. Differentiate numerator and denominator: $\lim_{x\rightarrow0}\frac{1-\cos x}{3x^{2}+2x}$

Step2: Apply L'Hopital's Rule again

Still in $\frac{0}{0}$ form. Differentiate numerator and denominator: $\lim_{x\rightarrow0}\frac{\sin x}{6x + 2}$

Step3: Substitute $x = 0$

Substitute $x = 0$ into $\frac{\sin x}{6x + 2}$: $\frac{\sin0}{6\times0+2}=\frac{0}{2}=0$

Answer:

$0$