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Question

Mathematics Question on Limits

If a function f(x)f(x) is given by f(x)=x1+x+x(x+1)(2x+1)+x(2x+1)(3x+1)+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=0x =0, f(x)f(x)

A

has no limit

B

is not continuous

C

is continuous but not differentiable

D

is differentiable

Answer

is not continuous

Explanation

Solution

Let f(x)=x1+x+x(x+1)(2x+1)+x(2x+1)(3x+1)+f(x)=\frac{x}{1+x}+\frac{x}{(x+1)(2 x+1)}+\frac{x}{(2 x+1)(3 x+1)}+\ldots \infty
=limnr=1nx[(r1)x+1](rx+1)=\displaystyle\lim _{n \rightarrow \infty} \displaystyle\sum_{r=1}^{n} \frac{x}{[(r-1) x+1](r x+1)}
=limxr=1n[x[(r1)x+1]1rx+1]=\displaystyle\lim _{x \rightarrow \infty} \displaystyle\sum_{r=1}^{n}\left[\frac{x}{[(r-1) x+1]}-\frac{1}{r x+1}\right]
=limn[11nx+1]=1=\displaystyle\lim _{n \rightarrow \infty}\left[1-\frac{1}{n x+1}\right]=1
For x=0x =0, we have f(x)=0f ( x )=0
Thus we have
f(x)={1,x0 0,x=0f(x) = \begin{cases} 1, & x \neq 0\\\ 0, & x = 0 \end{cases}
Clearly, limx0f(x)\displaystyle\lim _{x \rightarrow 0^{-}} f(x)
=limx0+f(x)f(0)=\displaystyle\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)
So, f(x)f (x) is not continuous at x=0x =0