Question Number 145779 by Engr_Jidda last updated on 08/Jul/21
$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({e}^{{t}} +{e}^{−{t}} −\mathrm{2}\right)\frac{{dt}}{\mathrm{1}−{cosx}} \\ $$
Answered by ArielVyny last updated on 08/Jul/21
![e^t +e^(−t) =2ch(t) (1/(1−cosx))∫_0 ^1 2ch(t)−2dt=(1/(1−cosx))[2sh(t)−2t]_0 ^1 f(x)=((e^1 +e^(−1) −2)/(1−cosx)) limf(x)_(x→0) =+∞](https://www.tinkutara.com/question/Q145786.png)
$${e}^{{t}} +{e}^{−{t}} =\mathrm{2}{ch}\left({t}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{cosx}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{ch}\left({t}\right)−\mathrm{2}{dt}=\frac{\mathrm{1}}{\mathrm{1}−{cosx}}\left[\mathrm{2}{sh}\left({t}\right)−\mathrm{2}{t}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${f}\left({x}\right)=\frac{{e}^{\mathrm{1}} +{e}^{−\mathrm{1}} −\mathrm{2}}{\mathrm{1}−{cosx}} \\ $$$${limf}\left({x}\underset{{x}\rightarrow\mathrm{0}} {\right)}=+\infty \\ $$