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  • If  <img alt="\mathrm{f(x)=\lim _{n \rightarrow \infty} \frac{\log (x+2)-x^{2 n} \sin x}{x^{2 n}+1} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Clim%20_%7B

If  \mathrm{f(x)=\lim _{n \rightarrow \infty} \frac{\log (x+2)-x^{2 n} \sin x}{x^{2 n}+1} }, then \mathrm{ f(x)} is

Option: 1

\mathrm{\text { continuous at } x= \pm 1}


Option: 2

\mathrm{\text { discontinuous at } x= \pm 1}


Option: 3

\mathrm{\text { continuous at } x= \pm 2}


Option: 4

\mathrm{\text { discontinuous at } x= \pm 2}


Answers (1)

best_answer

\mathrm{\text { If }|x|<1 \text {, then } f(x)=\lim _{x \rightarrow \infty} \frac{\log (x+2)-x^{2 n} \sin x}{x^{2 n}+1}}

                                            \mathrm{=\frac{\log (x+2)}{1}=\log (x+2)}

\mathrm{\text { If }|x|>1 \text {, then } f(x)=\lim _{n \rightarrow-} \frac{\log (x+2)-x^{2 n} \sin x}{x^{2 n}+1}}

                                          \mathrm{=\log _{x \rightarrow-\infty} \frac{\frac{\log (x+2)}{x^{2 n}}-\sin x}{1+\frac{1}{x^{2 n}}}=-\sin x}

\mathrm{\text { If }|x|=0 \text {, then } f(x)=\lim _{n \rightarrow \infty} \frac{\log (x+2)-x^{2 n} \sin x}{x^{2 n}+1}}

                                          \mathrm{=\frac{\log (x+2)-\sin x}{2}}

\mathrm{\text { Thus, } f(x)=\left\{\begin{array}{cc} -\sin x, & x<-1 \\ \log (x+2), & -1<x<1 \\ -\sin x, & x>1 \\ \frac{\log (x+2)-\sin x}{2}, & x= \pm 1 \end{array}\right.}

\mathrm{\text { Hence, } f(x) \text { is discontinuous at } x= \pm 1 \text {. }}

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