<s>[INST] How to prove this limit using epsilon delta method? i know how to construct a proof using epsilon delta method but as much as i have done i have to just algebraically manipulate and i got the value but i cant get this one.
$$\lim_{x \to c} \frac{x}{1+x^2} = \frac{c}{1+c^2}$$
My try: $$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon\implies\left|\frac{x^2-c^2}{1+c^2+x^2+x^2c^2}\right|<\epsilon\implies\left|\frac{(x-c)(x+c)}{1+c^2+x^2+x^2c^2}\right|<\epsilon$$
But i dont how to proceed after this?
 [/INST] If $c=0,$ take $\delta = \epsilon.$ If $c\ne 0,$ take $\delta = \min (\epsilon/(1+2c^2),|c|).$ Why? We have
$$\left |\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)} \right | \le |x-c|\,|1-xc|.$$
If $c=0,$ then we're left with $|x-c|$ on the right, perfect. If $c\ne 0,$ $|x-c|<|c| \implies |x|\le |x-c|+|c|<2|c|\implies |1-xc| < 1+2c^2.$ As $|x-c|<\epsilon/(1+2c^2),$ we're done.
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