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Question Number 160793 by cortano last updated on 06/Dec/21
    lim_(x→0)  ((2^(cos x)  − 2)/x^2 ) =?
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{\mathrm{cos}\:\mathrm{x}} \:−\:\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$
Answered by qaz last updated on 06/Dec/21
lim_(x→0) ((2^(cos x) −2)/x^2 )=lim_(x→0,ξ→1) (((cos x−1)2^ξ ln2)/x^2 )=−ln2
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{\mathrm{cos}\:\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{x}\rightarrow\mathrm{0},\xi\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)\mathrm{2}^{\xi} \mathrm{ln2}}{\mathrm{x}^{\mathrm{2}} }=−\mathrm{ln2} \\ $$
Answered by mr W last updated on 06/Dec/21
=lim_(x→0) ((−sin x 2^(cos x)  ln 2)/(2x))  =−((2×ln 2)/2)  =−ln 2
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{sin}\:{x}\:\mathrm{2}^{\mathrm{cos}\:{x}} \:\mathrm{ln}\:\mathrm{2}}{\mathrm{2}{x}} \\ $$$$=−\frac{\mathrm{2}×\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$=−\mathrm{ln}\:\mathrm{2} \\ $$

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