lim _{x} {rightarrow 0^{+}} frac{tan left(5(x | vedclass

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Question

(C) We use the standard limits: $\lim_{u ightarrow 0} \frac{	an u}{u} = 1$,$\lim_{u ightarrow 0} \frac{\ln(1+u)}{u} = 1$,$\lim_{u ightarrow 0}

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) We use the standard limits: $\lim_{u ightarrow 0} \frac{	an u}{u} = 1$,$\lim_{u ightarrow 0} \frac{\ln(1+u)}{u} = 1$,$\lim_{u ightarrow 0} \frac{	an^{-1} u}{u} = 1$,and $\lim_{u ightarrow 0} \frac{e^u-1}{u} = 1$. Given expression: $L = \lim _{x}$ ${ightarrow 0^{+}} \frac{	an \left(5 x^{1 / 3}ight)}{5 x^{1 / 3}} \cdot \frac{5 x^{1 / 3}}{\left(	an ^{-1} 3 x^{1 / 2}ight)^2} \cdot \frac{\ln(1+3 x^2)}{3 x^2} \cdot \frac{3 x^2}{e^{5 x^{4 / 3}}-1}$. Note that $\left(	an ^{-1} 3 x^{1 / 2}ight)^2 = (3 x^{1 / 2})^2 \cdot \left(\frac{	an ^{-1} 3 x^{1 / 2}}{3 x^{1 / 2}}ight)^2 = 9x \cdot (1)^2 = 9x$. Also,$e^{5 x^{4 / 3}}-1 = 5 x^{4 / 3} \cdot \frac{e^{5 x^{4 / 3}}-1}{5 x^{4 / 3}} = 5 x^{4 / 3} \cdot (1)$. Substituting these: $L = 1 \cdot \frac{5 x^{1 / 3}}{9x} \cdot 1 \cdot \frac{3 x^2}{5 x^{4 / 3}} = \frac{15 x^{7 / 3}}{45 x^{7 / 3}} = \frac{15}{45} = \frac{1}{3}$.

$\lim _{x}$ ${\rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\frac{4}{3}}}-1\right)}$ is equal to

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