How to answer the question from Calculus by Michael Spivak Chapter 5 Problem 14 
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*Prove that if $\lim\limits_{x\rightarrow0}{\frac{f(x)}x}=l$ and $b\neq 0$, then $\lim\limits_{x\rightarrow0}{\frac{f(bx)}x}=bl$. Hint: Write $\frac{f(bx)}x=b\frac{f(bx)}{bx}$

*What happens if $b=0$?

*Part 1 enables us to find $\lim\limits_{x\rightarrow0}{\frac{\sin{2x}}{x}}$ in terms of $\lim\limits_{x\to0}\frac{\sin x}x$. Find this limit in another way.


This is a question from Calculus by Michael Spivak Chapter 5 Problem 14.
 A: I was writing this as a comment, but that you-are-out-of-characters alert was on my nerve, so here it is:

Basically I'm new to proofs so I don't know where to start, can you give me a hint on how >to start a question regarding limits?

That's okay. First, learn basics and keep them in your mind. For example, in this question it's necessary to know that if $x \to 0$, then $bx \to 0$ if $b$ is a real nonzero constant. After that, try to rephrase the problem and use the hints. In this case you just need to understand there's no difference between $x$ and $bx$ when $x$ goes to zero. That is, let $g(x) = \frac{f(x)}{x}$. Then what is $g(bx)$? Assume $\lim_{x\to 0} g(x)=l$. Then it is clear that you can change $x$ into $bx$ and get $\lim_{bx\to 0} g(bx)=l$ Note that we haven't used the phrase "$x\to 0$ implies $bx \to 0$" yet, we are just putting a different value in $g$. To understand this, as in my comment, let $bx = y$ and $\lim_{bx\to 0} g(bx)=l$ becomes $\lim_{y\to 0} g(y)=l$, which is the same as our first assumption ($\lim_{x\to 0} g(x)=l$).
So we have $\lim_{bx\to 0} g(bx)=l$. Now use "$x\to 0$ implies $bx \to 0$". Thus $\lim_{x\to 0} g(bx)=l$. So, using the definition of $g$, we have $\lim_{x\to 0} \frac{f(bx)}{bx}=l$, which means $\lim_{x\to 0} \frac{f(bx)}{x}=bl$.
For $b=0$, the fraction $\frac{f(bx)}{bx}$ is not defined, and so the limit doesn't exist.
For part (c), set $g(x)=\sin x$.
A: (a) Suppose that $\lim_{x\to 0}f(x)/x=L$. We claim that $\lim_{x\to 0}f(bx)/x=bL$ where $b\not=0$.
$$\left|\frac{f(bx)}{x}-bL\right|=|b|\left|\frac{f(bx)}{bx}-L\right| \tag{1}$$
Given $\varepsilon>0$, there is a $\delta>0$ such that $|f(y)/y-L|<\varepsilon/|b|$ whenever $0<|y|<\delta$. Let $0<|x|<\delta/|b|$ (remember that $b\not=0$). Thus $0<|bx|<\delta$ and so $ |f(bx)/bx-L|< \varepsilon/|b|$, i.e., $|b||f(bx)/bx-L|< \varepsilon$ which by (1) is what we wanted to prove..
(b) $\lim_0f(0)/x$ may or may not exists, exists for example when $f(0)=0$.
(c) $$\frac{\sin2x}{x}=\frac{2 \sin x \cos x}{x}=2 \cos x\frac{\sin x}{x}\to 2$$
A: (a):
As in the hint 
$$\lim_{x\rightarrow 0} \frac{f(bx)}{x}=\lim_{x\rightarrow 0} b\frac{f(bx)}{bx}=b \lim_{y/b\rightarrow 0}  \frac{f(y)}{y}=b \lim_{y\rightarrow 0}  \frac{f(y)}{y}=bl$$
(b):
If $b=0$, then $\frac{f(bx)}{x}=\frac{f(0)}{x}$, so the limit as $x\rightarrow 0$ doesn't exist.
(c): Indeed we can use (a): Setting $f(x)=\sin(x)$, $b=2$ and $l=\lim_{x\rightarrow 0} \frac{\sin(x)}{x}$ (you might already know that $l=1$, one can show that by L'Hôspital's rule), we see
$$\lim_{x\rightarrow 0} \frac{\sin(2x)}{x} = 2 l=2$$
Another way to find this limit is directly by L'Hôspital:
$$\lim_{x\rightarrow 0} \frac{\sin(2x)}{x}=\lim_{x\rightarrow 0} \frac{2 \cos(2x)}{1}=2$$
