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Question Number 145129 by liberty last updated on 02/Jul/21

Answered by EDWIN88 last updated on 03/Jul/21

$$\underset{x\to 0^{+}}{\lim }\frac{\sin {}^2(\frac{1}{\sqrt{1-x^2}}-1)}{\cos {}^2(\frac{1}{\sqrt{1-x^2}}-1){\left({2\sin }^2\left(\frac{\sqrt{2x}}{2}\right)\right)}^n}=a$$$$\Rightarrow 1\times \frac{1}{2}\times {\left(\underset{x\to 0^{+}}{\lim }\frac{\sin (\frac{1}{\sqrt{1-x^2}}-1)}{\sin {}^n\left(\sqrt{\frac{x}{2}}\right)}\right)}^2=a$$$$\Rightarrow \frac{1}{2}\times {\left(\underset{x\to 0^{+}}{\lim }\frac{\cos (\frac{1}{\sqrt{1-x^2}}-1).x{(1-x^2)}^{-3/2}}{n\sin {}^{n-1}\left(\sqrt{\frac{x}{2}}\right)\cos \left(\sqrt{\frac{x}{2}}\right).\frac{1}{2\sqrt{2x}}}\right)}^2=a$$$$\Rightarrow \frac{1}{2}.\frac{1}{n^2}.\underset{x\to 0^{+}}{\lim }{\left(\frac{2x\sqrt{2x}}{\sin {}^{n-1}\left(\sqrt{\frac{x}{2}}\right)}\right)}^2=a$$$$\Rightarrow \frac{4}{n^2}.\underset{x\to 0^{+}}{\lim }\frac{x^3}{\sin {}^{2n-2}\left(\sqrt{\frac{x}{2}}\right)}=a$$$$\mathrm{let}\sqrt{x}=t\mathrm{\&}t\to 0$$$$\Rightarrow \frac{4}{n^2}.\underset{t\to 0}{\lim }\frac{t^6}{\sin {}^{2n-2}\left(\frac{t}{\sqrt{2}}\right)}=a$$$$sincelimitfinite,itfollows$$$$that2n-2=6orn=4$$$$sowegetlimitbecomes$$$$\Rightarrow \frac{1}{4}.\underset{t\to 0}{\lim }{\left[\frac{t}{\sin \left(\frac{t}{\sqrt{2}}\right)}\right]}^6=a$$$$\Rightarrow \frac{1}{4}\times {\left(\sqrt{2}\right)}^6=a$$$$\Rightarrow a=2.Hencen+a=4+2=6$$

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