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Let $f$ such that $\lim_{x\to 0}f(x)=\infty$ and let $g(x)=\sin(\frac{1}{x})$. I know that $g$ does not have a limit at $x=0$, but what about $\lim_{x\rightarrow 0}(f(x)+g(x))$?


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up vote 3 down vote accepted

Write $$f(x)+g(x)=f(x) \left( 1+\frac{g(x)}{f(x)} \right)$$ and notice that the fraction tends to zero since the numerator is bounded and the denominator diverges.

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Since $\sin(x)$ is bounded for all $x$ then $\sin(1/x)$ is also bounded.

Hence the limit is infinity.

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Always limit is infinity considering your problem.

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A rather synthetic answer. – Siminore May 9 '12 at 11:44
The statement is true, but it's rather useless to the OP, since it gives no explanation at all. – Brian M. Scott May 9 '12 at 17:53

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