$\lim_{x \to 0}(f+g)$ if $\lim_{x \to 0}g$ does not exist

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 $\lim_{x\rightarrow 0}(f(x)+g(x))$?

Thanks!

-

3 Answers

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.

-

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

Hence the limit is infinity.

-

Always limit is infinity considering your problem.

-
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