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

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Question

(A) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(	an^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}}$.પ્રમાણિત

Vedclass · Accepted Answer

$3/5$

Vedclass · Answer

(A) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(	an^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}}$.પ્રમાણિત લક્ષોનો ઉપયોગ કરતા: $\sin(x^{1/3}) \approx x^{1/3}$,$\ln(1+3x) \approx 3x$,$	an^{-1}\sqrt{x} \approx \sqrt{x}$,અને $e^{5x^{1/3}}-1 \approx 5x^{1/3}$.કિંમતો મૂકતા: $\frac{x^{1/3} \cdot 3x}{(\sqrt{x})^2 \cdot 5x^{1/3}} = \frac{3x^{4/3}}{5x^{4/3}} = \frac{3}{5}$.

$\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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