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

Correct Answer is : is not continuous

is continuous but not differentiable

Correct Answer is : is not continuous

Test Index

BITSAT 2015 Paper

Section: Mathematics

Question : 86 of 150

Marks: +1, -0

\frac{x}{1+x} + \frac{x}{(x+1)(2x+1)}

\frac{x}{(2x+1)(3x+1)}

+ ... ∞ , then at x = 0 , f (x)

Solution:

Let

f (x) =

\frac{x}{1+x} + \frac{x}{(x+1)(2x+1)}

\frac{x}{(2xx+1)(3x+1)}

+ ... ∞

\lim\limits_{x\rightarrow\infty} \sum\limits_{r=1}^{n}

\frac{x}{[(r-1)x+1](rx+1)}

\lim\limits_{x\rightarrow\infty} \sum\limits_{r=1}^{n}

\left[ \frac{x}{[(r-1)x+1]} - \frac{1}{rx+1} \right]

\lim\limits_{n\rightarrow\infty}

\left[ 1 - \frac{1}{nx+1} \right]

= 1

For x = 0, we have f(x) = 0

Thus, we have f (x) =

\begin{cases} 1 & x \neq 0 \\ 0 & x = 0 \end{cases}

Clearly,

\lim\limits_{x\rightarrow 0^{-}}

f (x) =

\lim\limits_{x\rightarrow 0^{+}}

f (x) ≠ f (0)

So, f(x) is not continuous at x = 0.

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