lim _{x rightarrow 0} frac{(3^{2x}-sqrt{x+1}) | vedclass

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(B) ધારો કે $L = \lim _{x \rightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x}$.નિત્યસમ $1 - \cos 4x = 2 \sin^2 2x$ નો ઉપયોગ કરતા,$L = \li

Vedclass · Accepted Answer

$\frac{5}{16} \log \left(\frac{81}{e}\right)$

Vedclass · Answer

(B) ધારો કે $L = \lim _{x ightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x}$.નિત્યસમ $1 - \cos 4x = 2 \sin^2 2x$ નો ઉપયોગ કરતા,$L = \lim _{x ightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{2 \sin^2 2x}$.અંશ અને છેદને $x^2$ વડે ભાગતા અને પ્રમાણિત લક્ષ $\lim_{x 	o 0} \frac{\sin ax}{ax} = 1$ નો ઉપયોગ કરતા:$L = \lim _{x ightarrow 0} \left( \frac{3^{2x} - \sqrt{x+1}}{x} ight) \cdot \left( \frac{\sin 5x}{x} ight) \cdot \left( \frac{x^2}{2 \sin^2 2x} ight) = \lim _{x ightarrow 0} \left( \frac{3^{2x} - 1 - (\sqrt{x+1} - 1)}{x} ight) \cdot 5 \cdot \frac{1}{2(2)^2}$.$L = \frac{5}{8} \lim _{x ightarrow 0} \left( \frac{3^{2x} - 1}{x} - \frac{\sqrt{x+1} - 1}{x} ight)$.$\lim_{x 	o 0} \frac{a^x - 1}{x} = \ln a$ અને $\lim_{x 	o 0} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{2}$ નો ઉપયોગ કરતા:$L = \frac{5}{8} \left( 2 \ln 3 - \frac{1}{2} ight) = \frac{5}{16} (4 \ln 3 - 1) = \frac{5}{16} (\ln 81 - \ln e) = \frac{5}{16} \ln \left( \frac{81}{e} ight)$.

$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$

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