lim _{x rightarrow 0} frac{63^x-9^x-7^x+1}{sq | vedclass

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(B) Let $L = \lim _{x \rightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$.Factor the numerator: $63^x-9^x-7^x+1 = 9^x(7^x-1) - 1(7^x-1) =

Vedclass · Accepted Answer

$4 \sqrt{2} \log 7 \cdot \log 9$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$.Factor the numerator: $63^x-9^x-7^x+1 = 9^x(7^x-1) - 1(7^x-1) = (9^x-1)(7^x-1)$.Simplify the denominator: $\sqrt{2}-\sqrt{1+\cos x} = \sqrt{2}-\sqrt{2\cos^2(x/2)} = \sqrt{2}(1-\cos(x/2))$.Using the identity $1-\cos 	heta = 2\sin^2(	heta/2)$,we have $\sqrt{2}(2\sin^2(x/4)) = 2\sqrt{2}\sin^2(x/4)$.Thus,$L = \lim _{x ightarrow 0} \frac{(9^x-1)(7^x-1)}{2\sqrt{2}\sin^2(x/4)}$.Using the standard limit $\lim _{x ightarrow 0} \frac{a^x-1}{x} = \ln a$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$:$L = \lim _{x}$ ${ightarrow 0} \frac{(\frac{9^x-1}{x})(\frac{7^x-1}{x})x^2}{2\sqrt{2}(\frac{\sin(x/4)}{x/4})^2 (x/4)^2} = \frac{\ln 9 \cdot \ln 7}{2\sqrt{2} \cdot (1/16)} = \frac{16 \ln 9 \cdot \ln 7}{2\sqrt{2}} = 4\sqrt{2} \ln 9 \ln 7$.

$\lim _{x \rightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\ldots$.

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