what such cases exist?
Such that $\lim_{x\to a} f(x)g(x)$ exists even though neither $\lim_{x\to a} f(x)$ nor $\lim_{x\to a} g(x)$ exists.
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Do you really mean $x\to b$ in the $g(x)$ limit? – Thomas Andrews Aug 26 '14 at 02:50
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No no it's x->a – Danxe Aug 26 '14 at 03:24
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Suppose $f(x)=g(x)=1$ when $x$ is rational but $-1$ when $x$ is irrational. In this example you can let $a$ (the approached point) be anything, e.g. $a=0.$
coffeemath
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$$f(x)=1+\tan(x)^2$$ $$g(x)=\cos(x)^2$$ $$\text{ So that we have: } \text{ }\text{ } f(x)g(x)=\sin(x)^2+\cos(x)^2=1$$ $$\text{ But neither} \lim_{x\to \infty}f(x) \text{ nor} \lim_{x\to\infty} g(x) \text{ exist }$$ $$\text{ While}\lim_{x\to\infty}f(x)g(x)=1 \text{ does exist}$$
Ethan Splaver
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