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Let $f:\mathbb{R}^p \rightarrow \mathbb{R}^n$ and $g: \mathbb{R}^n \rightarrow \mathbb{R}^m$. Suppose that $\lim\limits_{x \to a} f(x)=L\lim\limits_{x \to L} g(x)=M$. Show that if $g$ is continuous at $L$ then $$\lim\limits_{x \to a} g(f(x))=M$$

Not sure how to get started here... Any hints will be appreciated!

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    I'm assuming $b = f(a)$. Otherwise it wouldn't be true – 2017-02-14

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Let $\epsilon>0$. Since $\lim_{x\rightarrow L}g(x)=M$, there exists a $\delta_1>0$ such that $$ |x-L|<\delta_1\Rightarrow |g(x)-M|<\epsilon $$

Since $\lim_{x\rightarrow a}f(x)=L$, there exists a $\delta_2>0$ such that $$ |x-a|<\delta_2\Rightarrow |f(x)-L|<\delta_1 $$ Then, $$ |x-a|<\delta_2\Rightarrow |f(x)-L|<\delta_1\Rightarrow |g(f(x))-M|<\epsilon. $$

We just proved that for all $\epsilon>0$ there exists a $0< \delta:=\delta_2$ such that $|x-a|<\delta\Rightarrow|g(f(x))-M|<\epsilon$, which is the definition of $$ \lim_{x\rightarrow a}g(f(x))=M. $$

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    Why is Since $\lim_{x\rightarrow a}f(x)=L$, there exists a $\delta_2>0$ such that $$ |x-a|<\delta_2\Rightarrow |f(x)-L|<\delta_1$$ the case? In other wors, why is the last part less than $\delta_1$ ? – 2017-02-23
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    Because, by the definition of limit, $\lim_{x\rightarrow a}f(x)=L$ is equivalent to say that, FOR ALL $\epsilon >0$, there exists a $\delta>0$ such that $|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$. I just chose $\epsilon:=\delta_1$ to make it work. – 2017-02-23