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Question Number 104667 by bobhans last updated on 23/Jul/20
lim_(x→0)  [(3^(πx) /(tan x)) − ((cos^2 x)/(sin x))] ?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{3}^{\pi{x}} }{\mathrm{tan}\:{x}}\:−\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}\right]\:? \\ $$
Answered by bramlex last updated on 23/Jul/20
lim_(x→0) ((3^(πx) .cos x)/(sin x))− ((cos^2 x)/(sin x)) =  lim_(x→0)  ((cos x)/(sin x)) {3^(πx) −cos x } =   lim_(x→0) ((x.cos x)/(sin x)) {((3^(πx) −cos x)/x)} =  lim_(x→0)  cos x. (x/(sin x)) {((3^(πx) −1+1−cos x)/x)}=  1 ×lim_(x→0)  {((3^(πx) −1)/x) + ((1−cos x)/x)} =  1×lim_(x→0)  {((π.ln (3).3^(πx) )/1) + 0 } =  π.ln (3) ★
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}^{\pi{x}} .\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}−\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:\left\{\mathrm{3}^{\pi{x}} −\mathrm{cos}\:{x}\:\right\}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}.\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:\left\{\frac{\mathrm{3}^{\pi{x}} −\mathrm{cos}\:{x}}{{x}}\right\}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\:{x}.\:\frac{{x}}{\mathrm{sin}\:{x}}\:\left\{\frac{\mathrm{3}^{\pi{x}} −\mathrm{1}+\mathrm{1}−\mathrm{cos}\:{x}}{{x}}\right\}= \\ $$$$\mathrm{1}\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left\{\frac{\mathrm{3}^{\pi{x}} −\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}}\right\}\:= \\ $$$$\mathrm{1}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left\{\frac{\pi.\mathrm{ln}\:\left(\mathrm{3}\right).\mathrm{3}^{\pi{x}} }{\mathrm{1}}\:+\:\mathrm{0}\:\right\}\:= \\ $$$$\pi.\mathrm{ln}\:\left(\mathrm{3}\right)\:\bigstar\: \\ $$
Commented by bobhans last updated on 23/Jul/20
very nice ! ⌣^o  !
$${very}\:{nice}\:!\:\overset{{o}} {\smile}\:! \\ $$

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