Lim x->c[f(x)+g(x)]: Examples to Show No Implication

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Can someone point me in the right dirrection with this problem:

find examples to show that if

a) lim x->c [f(x)+g(x)] exists, this does not imply that either lim x->c [f(x)] or lim x->c [g(x)] exists.

b) lim x->c [f(x)*g(x)] exists, this does not imply that either lim x->c [f(x)] or lim x->c [g(x)] exists.

any help would be great
 
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Think simply: if f(x)= 1/x, what is the limit as x-> 0? what is limit of g(x)= -1/x as x-> 0? What about f(x)+ g(x)?
 
would that still work for part B of the problem?
 
Not that example because there is no limit of f(x)*g(x). Use the two examples of x^2 and 1/x^2
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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