mathop {lim }limits_{x to 0} frac{{sin ({x^{1 | vedclass

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(A) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(	an^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)

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$3/5$

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(A) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(	an^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}}$.Using standard limits $\mathop {\lim }\limits_{u 	o 0} \frac{\sin u}{u} = 1$,$\mathop {\lim }\limits_{u 	o 0} \frac{\ln(1+u)}{u} = 1$,$\mathop {\lim }\limits_{u 	o 0} \frac{	an^{-1} u}{u} = 1$,and $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$.We rewrite the expression as:$\mathop {\lim }\limits_{x 	o 0} \left[ \frac{\sin(x^{1/3})}{x^{1/3}} \cdot \frac{\ln(1+3x)}{3x} \cdot 3x \cdot \frac{1}{(\frac{	an^{-1}\sqrt{x}}{\sqrt{x}})^2 \cdot x} \cdot \frac{5x^{1/3}}{e^{5x^{1/3}}-1} \cdot \frac{1}{5x^{1/3}} ight]$$= \mathop {\lim }\limits_{x 	o 0} \left[ 1 \cdot 1 \cdot 3x \cdot \frac{1}{1^2 \cdot x} \cdot 1 \cdot \frac{1}{5x^{1/3}} ight]$$= \mathop {\lim }\limits_{x 	o 0} \frac{3x}{5x^{4/3}} = \mathop {\lim }\limits_{x 	o 0} \frac{3}{5x^{1/3}}$.Wait,let us re-evaluate the powers. The expression is $\frac{\sin(x^{1/3}) \ln(1+3x)}{(	an^{-1}\sqrt{x})^2 (e^{5x^{1/3}}-1)}$.As $x 	o 0$,$\sin(x^{1/3}) \approx x^{1/3}$,$\ln(1+3x) \approx 3x$,$	an^{-1}\sqrt{x} \approx \sqrt{x}$,and $e^{5x^{1/3}}-1 \approx 5x^{1/3}$.Substituting these approximations: $\frac{x^{1/3} \cdot 3x}{(\sqrt{x})^2 \cdot 5x^{1/3}} = \frac{3x^{4/3}}{5x \cdot x^{1/3}} = \frac{3x^{4/3}}{5x^{4/3}} = \frac{3}{5}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(\tan^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}} = $

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