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If $F(x) = x^{2} + \frac{x^{2}}{1+ x^{2}} +\frac{x^{2}}{\left ( 1+ x^{2} \right)^{2}} +\dots+\frac{x^{2}}{\left ( 1+ x^{2} \right )^{n}} + \dots$

then at $x=0$, is $F(x)$ continuous or not?

My try :

I tried for both

$$\lim_{h \to 0^+} = (0+h)^{2} + \frac{(0+h)^{2}}{1+ (0+h)^{2}} +\frac{(0+h)^{2}}{\left ( 1+ (0+h)^{2} \right)^{2}} +\dots+\frac{(0+h)^{2}}{\left ( 1+ (0+h)^{2} \right )^{n}} + \dots$$

and similarly,

$$\lim_{h \to 0^-} = (0-h)^{2} + \frac{(0-h)^{2}}{1+ (0-h)^{2}} +\frac{(0-h)^{2}}{\left ( 1+ (0-h)^{2} \right)^{2}} +\dots+\frac{(0-h)^{2}}{\left ( 1+ (0-h)^{2} \right )^{n}} + \dots$$

I am getting this function continuous, but not sure though ...

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    The value of $h$ when $h$ is negative is $h$, not $-h$. – 2017-01-02
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    Did you compute those limits? Assuming you did and showed them equal, you showed that $\lim_{h \to 0} F(x)$ exists, and computed its value. But you seem to have forgotten to check whether $F(0) = \lim_{h \to 0} F(x)$. – 2017-01-02
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    @Hurkyl My bad !! I missed that. I got my error now. Thanks !! – 2017-01-02

1 Answers 1

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We have: $F(0) = 0$, and for $x \neq 0, F(x) = x^2\left(1+\dfrac{1}{1+x^2}+ \dfrac{1}{(1+x^2)^2} + \cdots \right)= x^2\left(\dfrac{1}{1-\dfrac{1}{1+x^2}}\right)=1+x^2$. We now check continuity of $F$ at $x = 0$. We have: $\displaystyle \lim_{x \to 0} F(x) = 1 \neq 0 = F(0)$, hence $F$ is not continuous at $x = 0$, but continuous at $x \neq 0$.