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Let $a\in\mathbb{R}$ and let $f$ and $g$ be real functions defined at all points $x$ in some open interval containing $a$ except possibly at $x=a$. Decide which of the following statements are true and which are false. Prove the true one and give counterexamples for the false one.

For each $n\in\mathbb{N}$, the fuction $\left( x-a\right) ^{n}\sin \left( f\left( x\right) \left( x-a\right) ^{-n}\right)$ has a limit as $x\rightarrow a$.

Can you give a hint?

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    Hint: $\sin$ is bounded in $\mathbb R$. – 2017-01-24

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Hint. We have that $|\sin (x)|\leq 1$ for all $x\in \mathbb R$. Hence

$$|(x-a)^n\sin(f(x)(x-a)^{-n})|=|(x-a)^n|\cdot|\sin(f(x)(x-a)^{-n})|\leq |(x-a)^n|.$$

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    Then, if $\left| x-a\right| < \delta$ then $\left| \left( x-a\right) ^{n}-L\right|$=... (I should do smaller than epsilon this, but how?) – 2017-01-24
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    Show that the limit is zero. Let $\delta = \min(1,\varepsilon)$ so $\delta \leq 1$ and $\delta \leq \varepsilon$. Hence, if $|x-a|<\delta$, then $$|(x-a)^n-0| = |x-a|^n < \delta^n \leq \delta \leq \varepsilon. $$ – 2017-01-24
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    If we pick $\delta=\varepsilon$ then is it enough? So, should we show $\delta^{n}\leq \delta$? – 2017-01-24
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    It is not quite enough, because if $\varepsilon > 1$ then it is not true that $\delta^n \leq \delta$ (in fact, it is larger). Hence I chose $\delta = \min(1,\varepsilon)$. – 2017-01-24