The value of lim _{x rightarrow 0} frac{15^{x | vedclass

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Question

(A) Given limit: $\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$ Factorizing the numerator: $15^{x}-5^{x}-3^{x}+1 = 5^{x}(3^{x}-1) -

Vedclass · Accepted Answer

$\frac{(\log 3)(\log 5)}{2}$

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(A) Given limit: $\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$ Factorizing the numerator: $15^{x}-5^{x}-3^{x}+1 = 5^{x}(3^{x}-1) - 1(3^{x}-1) = (3^{x}-1)(5^{x}-1)$ Using the identity $1-\cos 2x = 2\sin^{2}x$: $\lim _{x \rightarrow 0} \frac{(3^{x}-1)(5^{x}-1)}{2\sin^{2}x}$ Dividing numerator and denominator by $x^{2}$: $\lim _{x \rightarrow 0} \frac{(\frac{3^{x}-1}{x})(\frac{5^{x}-1}{x})}{2(\frac{\sin x}{x})^{2}}$ Using the standard limit $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x} = \log a$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$: $= \frac{(\log 3)(\log 5)}{2(1)^{2}} = \frac{(\log 3)(\log 5)}{2}$

The value of $\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$ is

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