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  • Let <img alt="\mathrm{f(x)=\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{x}{(r x+1)\{(r+1) x+1\}}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Clim%20_%7Bn%

Let \mathrm{f(x)=\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{x}{(r x+1)\{(r+1) x+1\}}}, then

Option: 1

\mathrm{f(x)} is continuous but not differentiable at \mathrm{x=0}


Option: 2

\mathrm{f(x)}  is both continuous and differentiable at \mathrm{x= 0}


Option: 3

\mathrm{f\left ( x \right )} is neither continuous nor differentiable at \mathrm{x= 0}


Option: 4

\mathrm{f\left ( x \right )} is a periodic function


Answers (1)

best_answer

\mathrm{t_{r+1}=\frac{x}{(r x+1)\{(r+1) x+1\}}=\frac{(r+1) x+1-(r x+1)}{(r x+1)[(r+1) x+1]}=\frac{1}{(r x+1)}-\frac{1}{(r+1) x+1}}

\mathrm{ \therefore \\ \quad S_n=\sum_{r=0}^{n-1} t_{r+1}=\left\{\begin{array}{c} 1-\frac{1}{n x+1}, x \neq 0 \\ 0 \quad, \quad x=0 \end{array}\right.}

\mathrm{\therefore \quad f(x)=\lim _{n \rightarrow \infty} S_n=\left\{\begin{array}{ll} 1, x \neq 0 \\ 0, x=0 \quad & \lim _{x \rightarrow 0} f(x)=1 \end{array} \text { and } f(0)=0\right.}

Hence \mathrm{f(x)} is neither continuous nor differentiable at \mathrm{x=0}.
Clearly \mathrm{f(x)} is not a periodic function

Posted by

Divya Prakash Singh

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