3
$\begingroup$
  1. 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 $\frac{f(bx)}x=b\frac{f(bx)}{bx}$
  2. What happens if $b=0$?
  3. 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.

$\endgroup$
  • $\begingroup$ @Amir Hossein 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? $\endgroup$ – shinobi20 Mar 28 '14 at 13:24
  • $\begingroup$ I've posted a new answer. Read it please. $\endgroup$ – Amir Hossein Mar 28 '14 at 14:49
  • $\begingroup$ It is generally discouraged to post questions are images. If you are new to latex there are some online tools that helps you. See code-tags or daum equation editor:google chrome app or mathjax basic tutorial and quick reference. $\endgroup$ – jdoicj Mar 30 '14 at 8:35
2
$\begingroup$

(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) $$\frac{\sin2x}{x}=\frac{2 \sin x \cos x}{x}=2 \cos x\frac{\sin x}{x}\to 2$$

$\endgroup$
1
$\begingroup$

(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:

$$\lim_{x\rightarrow 0} \frac{\sin(2x)}{x}=\lim_{x\rightarrow 0} \frac{2 \cos(2x)}{1}=2$$

$\endgroup$
1
$\begingroup$

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 $g(x)=\sin x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.