If a function f(x) is given by f(x)=(x)/(1+x)+(x)/((x+1)(2 x+1))+(x)/((2 x+1)(3 x+1))+ , ∞ , then at x=0, f(x)

Correct answer: B. is not continuous Let f(x)=(x)/(1+x)+(x)/((x+1)(2 x+1))+(x)/((2 x+1)(3 x+1))+ ∞= _ n arrow ∞ ∑_ r=1 ^ n (x)/([(r-1) x+1](r x+1))= _ x arrow ∞ ∑_ r=1 ^ n [ x [(r-1) x+1] - 1 rx+1 ]= _ n arrow ∞ [1-(1)/(n x+1)]=1 For x=0, we have f(x)=0 Thus, we have f(x)=\ array ll 1, & x ≠ 0 0, & x=0 array . Clearly, _ x arrow 0^ - f(x)= _ x arrow 0^ + f(x) ≠ f(0) So, f(x) is not continuous at x=0 .

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Continuity and Differentiability

BITSAT 2015 · Mathematics · Continuity and Differentiability

If a function f(x) is given by f(x)=(x)/(1+x)+(x)/((x+1)(2...

If a function

f(x)

is given by

f(x)=\frac{x}{1+x}+\frac{x}{(x+1)(2 x+1)}+\frac{x}{(2 x+1)(3 x+1)}+\ldots, \infty

, then at

x=0, f(x)

Options

Ahas no limit
Bis not continuous
Cis continuous but not differentiable
Dis differentiable

Correct answer

B. is not continuous

Step-by-step solution

Let

f(x)=\frac{x}{1+x}+\frac{x}{(x+1)(2 x+1)}+\frac{x}{(2 x+1)(3 x+1)}+\ldots \infty=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{x}{[(r-1) x+1](r x+1)}=\lim _{x \rightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}}

\left[\frac{\mathrm{x}}{[(\mathrm{r}-1) \mathrm{x}+1]}-\frac{1}{\mathrm{rx}+1}\right]=\lim _{n \rightarrow \infty}

\left[1-\frac{1}{n x+1}\right]=1

For

x=0,

we have

f(x)=0

Thus, we have

f(x)=\left\{

\begin{array}{ll}1, & x \neq 0 \\ 0, & x=0\end{array}

\right.

Clearly,

\lim _{x \rightarrow 0^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x}) \neq \mathrm{f}(0)

So,

\mathrm{f}(\mathrm{x})

is not continuous at

\mathrm{x}=0

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