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  • <img alt="\mathrm{ f(x)=\lim _{n \rightarrow \infty} \frac{\tan \pi x^2+(x+1)^n \sin x}{x^2+(x+1)^n} \text {, find } \lim _{x \rightarrow 0} f(x) \text {. }}" src="https://entrancecorner.oncodecogs

\mathrm{ f(x)=\lim _{n \rightarrow \infty} \frac{\tan \pi x^2+(x+1)^n \sin x}{x^2+(x+1)^n} \text {, find } \lim _{x \rightarrow 0} f(x) \text {. }}

Option: 1

limit does not exist


Option: 2

limit exist, limit=1


Option: 3

limit exist,limit=-1


Option: 4

none


Answers (1)

best_answer

\mathrm{f(x)=\lim _{n \rightarrow \infty} \frac{\tan \pi x^2+(x+1)^n \sin x}{x^2+(x+1)^n}}\\ \mathrm{ if ~x>0, so 1+x>1 so, (1+x)^n \rightarrow \infty ~when ~n \rightarrow \infty }
\begin{aligned} &\mathrm{ f(x)=\lim _{n \rightarrow \infty} \frac{\frac{\tan \left(\pi x^2\right)}{(x+1)^n}+\sin x}{\frac{x^2}{(x+1)^n}+1} \text { when } x>0 }\\ & \mathrm{=\sin x} \\ & \text { If }\mathrm{ x<0, \text { so, } 1+x<1 \operatorname{so}(1+x)^n \rightarrow 0 \text { when } n \rightarrow \infty} \\ &\mathrm{ f(x)=\lim _{n \rightarrow \infty} \frac{\tan \left(\pi x^2\right)+(x+1)^n \sin x}{x^2+(x+1)^n}=\frac{\tan \pi x^2}{x^2} . }\\ & \mathrm{f(x)}=\left\{\begin{array}{l} \mathrm{\sin x \quad x>0 }\\ \mathrm{\frac{\tan \left(\pi x^2\right)}{x^2}, x<0} \end{array}\right. \\ & \mathrm{\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} \sin x=0 }\\ &\mathrm{ \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi x^2\right)}{x^2}=\pi }\end{aligned}
Thus, limit does not exist.

 

Posted by

shivangi.shekhar

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