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  • If <img alt="\mathrm{f(x)=\lim _{n \rightarrow \infty} \frac{\left(1-\cos \left(1-\tan \left(\frac{\pi}{4}-x\right)\right)\right)(x+1)^n+\lambda \sin \left(\left(n-\sqrt{n^2-8 n}\right) x

If

\mathrm{f(x)=\lim _{n \rightarrow \infty} \frac{\left(1-\cos \left(1-\tan \left(\frac{\pi}{4}-x\right)\right)\right)(x+1)^n+\lambda \sin \left(\left(n-\sqrt{n^2-8 n}\right) x\right)}{x^2(x+1)^n+x}},

\mathrm{x\neq 0} is continuous at \mathrm{x= 0}, then find the value of \mathrm{(f(0)+2 \lambda)}

Option: 1

3


Option: 2

2


Option: 3

1


Option: 4

0


Answers (1)

1. For right hand limit, as  \mathrm{n\rightarrow \infty} , and  \mathrm{x \rightarrow 0^{+}}, note that \mathrm{\sin \left(\frac{8 n x}{n+\sqrt{n^2-8 n}}\right) \rightarrow 4 x \rightarrow 0}  . But

\mathrm{(x+1)^n \rightarrow \infty}. So the limit is of form \mathrm{\infty/\infty} . Divide by \mathrm{x^2(x+1)^n}  on numerator and

denominator and the limit is 

                                  \mathrm{\begin{gathered} \lim _{x \rightarrow 0^{+}, n \rightarrow \infty} \frac{\frac{2 \sin ^2\left(\frac{\cos (x)}{1+\tan (x)}\right)(x+1)^n}{x^2(x+1)^n}+\frac{\sin \left(\frac{\operatorname{sen} x}{\sin \sqrt{x^2-x^n}}\right)}{x^2(1+x)^n}}{1+\frac{x}{x^2(1+x)^n}} \\\\ =\frac{\lim _{x \rightarrow 0^{+}, n \rightarrow \infty} \frac{2 \sin ^2\left(\frac{\tan (x)}{1+\cos (x)}\right)(x+1)^n}{x^2(x+1)^n}+0}{1+0} \\\\ =\lim _{x \rightarrow 0^{+}} \frac{2 \sin ^2\left(\frac{\tan (x)}{1+\tan (x)}\right)}{x^2}=2 \end{gathered}}

2. For left limit, as \mathrm{n \rightarrow \infty} and \mathrm{x \rightarrow 0^{-},(x+1)^n \rightarrow 0} , so limit is of form 0/0:

                               \mathrm{\begin{gathered} \lim _{x \rightarrow 0, n \rightarrow \infty} \frac{2 \sin ^2\left(\frac{\tan (x)}{1+\tan (x)}\right)(x+1)^n+\lambda \sin \left(\frac{8 n x}{n+\sqrt{n^2-8 n}}\right)}{x^2(x+1)^n+x} \\\\ =\lim _{x \rightarrow 0^{-}, n \rightarrow \infty} \frac{2 \sin ^2\left(\frac{\tan (x)}{1+\tan (x)}\right) \frac{(x+1)^n}{x}+\frac{\lambda}{x} \sin \left(\frac{8 n x}{n+\sqrt{n^2-8 n}}\right)}{x(x+1)^n+1} \\\\ =\frac{0+4 \lambda}{0+1} \end{gathered}}

For continuity we need \mathrm{4 \lambda=2=f(0) \text { or } \lambda=1 / 2 \text {. So } f(0)+2 \lambda=2+1=3} .  

Posted by

Ramraj Saini

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