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Question Number 74015 by mathmax by abdo last updated on 17/Nov/19

$$letf\left(x\right)={\int }_x^{x^2}\frac{sh\left(xt\right)}{sin\left(xt\right)}dt$$$$calculate{lim}_{x\to 0}f\left(x\right)$$

Commented by mathmax by abdo last updated on 18/Nov/19

$$f\left(x\right)={\int }_x^{x^2}\frac{sh\left(xt\right)}{sin\left(xt\right)}dt{=}_{xt=u}\frac{1}{x}{\int }_{x^2}^{x^3}\frac{sh\left(u\right)}{sin\left(u\right)}du$$$$\mathrm{\exists }c\in ]x^2,x^3[/{\int }_{x^2}^{x^3}\frac{sh\left(u\right)}{sin\left(u\right)}du=sh\left(c\right){\int }_{x^2}^{x^3}\frac{du}{sinu}$$$${\int }_{x^2}^{x^3}\frac{du}{sinu}{=}_{tan\left(\frac{u}{2}\right)=t}{\int }_{tan\left(\frac{x^2}{2}\right)}^{tan\left(\frac{x^3}{2}\right)}\frac{1}{\frac{2t}{1+t^2}}\times \frac{2dt}{1+t^2}={[ln\mid t\mid ]}_{tan\left(\frac{x^2}{2}\right)}^{tan\left(\frac{x3}{2}\right)}$$$$=ln\mid \frac{tan\left(\frac{x^3}{2}\right)}{tan\left(\frac{x^2}{2}\right)}\mid \sim ln\mid x\mid (x\to 0)$$$$c=\lambda x^2+(1-\lambda )x^{3}with0<\lambda <1\Rightarrow f\left(x\right)\sim \frac{sh\left(c\right)}{x}ln\mid x\mid$$$$\sim sh(\lambda x^2+(1-\lambda )x^3)ln\mid x\mid \sim (\lambda x^2+(1-\lambda )x^3)ln\mid x\mid$$$$=x^{2}ln\mid x\mid (\lambda +(1-\lambda )x)\to 0(x\to 0)\Rightarrow {lim}_{x\to 0}f\left(x\right)=0$$$$$$

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