Question : Evaluate the following limit \[\lim_{x\rightarrow\frac{\pi}{2}\ }{\frac{\sin{\left(2x\right)}}{x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}}}\] Solution : To evaluate the limit \[\lim_{x\rightarrow\frac{\pi}{2}\ }{\frac{\sin{\left(2x\right)}}{x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}}}\] Since \[sin(2\ast\frac{\pi}{2})=sin(\pi)=0 \] and \[\left(\frac{\pi}{2}\right)^2+\frac{\pi}{2}\ast\frac{\pi}{2}-\frac{\pi^2}{2}=\frac{\pi^2}{4}+\frac{\pi^2}{4}-\frac{\pi^2}{2}=\frac{\pi^2}{2}-\frac{\pi^2}{2}=0\] That is \[\lim_{x\rightarrow\frac{\pi}{2}\ }{\frac{\sin{\left(2x\right)}}{x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}}} \] Is in \[\frac{0}{0}\ form\] Hence applying L Hospital rule \[\lim_{x\rightarrow\frac{\pi}{2}\ }{\frac{\sin{\left(2x\right)}}{x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}}}\]\[=\lim_{x\rightarrow0}{\frac{\frac{d}{dx}\left(\sin{\left(2x\right)}\right)}{\frac{d}{dx}\left(x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}\right)}}\] \[=\lim_{x\rightarrow0}{\frac{2\cos{\left(2x\right)}}{2x+\frac{\pi}{2}}}\] \[=\frac{2\ast\cos{\left(2\ast\frac{\pi}{2}\right)}}{2\ast\frac{\pi}{2}+\frac{\pi}{2}} =\frac{2\ast\left(-1\right)}{\frac{3\pi}{2}}\]…
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