# Finding a limit using the squeeze theorem

I've got this problem:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $ax^2 - 5x^4 \leq f(x) \leq ax^2$, $a \neq 0$

1) Find $\lim_{x \to 0}f(x)$

2) Find $a$ so that $\lim_{x \to 0}\frac{f(x)}{sin^2(x)} = 3$

1) By the squeeze theorem, I've got: $\lim_{x \to 0}ax^2 - 5x^4 = 0$ and $lim_{x \to 0}ax^2 = 0$, so $lim_{x\to0}f(x)=0$

2) Here I'm not sure if I'm doing things right. I've tried by using the squeeze theorem again in this way: $\frac{ax^2 - 5x^4}{sin^2(x)} \leq \frac{f(x)}{sin^2(x)} \leq \frac{ax^2}{sin^2(x)}$

Now I take these limits:

$\lim_{x\to0}\frac{ax^2}{sin^2(x)} = 3$ and $\lim_{x\to0}\frac{ax^2}{sin^2(x)}=3$ and here is where I'm stuck on since I can't figure out how to find $a$.

Is it ok to do, for example:

$\frac{ax^2}{sin^2(x)}=3 \to ax^2=3sin^2(x) \to a=3\frac{sin^2(x)}{x^2}$, take the limit $\lim_{x\to0}3\frac{sin^2(x)}{x^2} = 3$, so $a=3$

And:

$\frac{ax^2-5x^4}{sin^2(x)}=3 \to ax^2-5x^4=3sin^2(x) \to ax^2=3sin^2(x)+5x^4 \to a=3\frac{sin^2(x)}{x^2} + 5x^2$, then take the limit $\lim_{x\to0}3\frac{sin^2(x)}{x^2} + 5x^2 = 3$, so $a=3$ and then by the squeeze theorem, $\lim_{x\to0}\frac{f(x)}{sin^2(x)} = 3$

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Lucas: Just for clarification: is $sen$ the sine function? I've seen this before, so I'm thinking it's a language difference? –  amWhy Jan 28 '13 at 15:16
@amWhy I'm sorry, yes, it's the sine function. I've fixed the question. Thanks! –  Lucas Jan 28 '13 at 15:21

You want to find $a$ so that $$\lim_{x\to 0}\frac{ax^2}{\sin^2x}=3\;;\tag{1}$$ recall that $\lim_{x\to 0}\frac{\sin x}x=1$, and write

$$3=\lim_{x\to 0}\frac{ax^2}{\sin^2x}=a\lim_{x\to 0}\frac{x^2}{\sin^2x}=a\left(\lim_{x\to 0}\frac{x}{\sin x}\right)^2=a(1)^2=a\;.$$

You asked whether this was okay:

$\frac{ax^2}{sin^2(x)}=3 \to ax^2=3sin^2(x) \to a=3\frac{sin^2(x)}{x^2}$, take the limit $\lim_{x\to0}3\frac{sin^2(x)}{x^2} = 3$, so $a=3$

It simply isn’t true that $\frac{ax^2}{\sin^2 x}=3$: the limit on the lefthand side is essential. This means that you can’t perform the algebraic manipulations that you used. You could argue that if $(1)$ holds, then clearly $a\ne 0$, so dividing by $3a$ yields $$\frac1a=\lim_{x\to 0}\frac{x^2}{3\sin^2 x}=\frac13\left(\lim_{x\to 0}\frac{x}{\sin x}\right)^2=\frac13\;,$$ because $a$ and $3$ are constants and can be moved in and out of the limit.

What’s really going on here is a little more complicated, though: you don’t actually know at this point that you want to solve $(1)$ for $a$.

You know that $$ax^2-5x^4\le f(x)\le ax^2\;,$$ and for $x$ close to but not equal to $0$ you know that $\sin^2x>0$. Those are the only $x$ of interest when we compute a limit as $x\to 0$, so we may divide through by $\sin^2x$ to get

$$\frac{ax^2-5x^4}{\sin^2x}\le\frac{f(x)}{\sin^2x}\le\frac{ax^2}{\sin^2x}\;.$$

In the limit as $x\to 0$ we want the middle term to be $3$, so we must have

$$\lim_{x\to 0}\frac{ax^2-5x^4}{\sin^2x}\le3\le\lim_{x\to 0}\frac{ax^2}{\sin^2x}\;.\tag{2}$$

As we’ve seen, $$\lim_{x\to 0}\frac{ax^2}{\sin^2x}=a\lim_{x\to 0}\left(\frac{x}{\sin x}\right)^2=a(1)^2=a\;,$$ so $(2)$ requires that $a\ge 3$. On the other hand,

$$\lim_{x\to 0}\frac{ax^2-5x^4}{\sin^2x}=\lim_{x\to 0}\frac{ax^2}{\sin^2x}-5\lim_{x\to 0}x^2\left(\frac{x}{\sin x}\right)^2=a-5(0)(1)^2=a\;,$$

so $(2)$ also requires that $a\le 3$, and it follows that $a=3$.

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Thank you Brian, this really helps me clarifying things out! –  Lucas Jan 29 '13 at 12:31
@Lucas: You’re welcome; I’m glad it helped. –  Brian M. Scott Jan 29 '13 at 18:26

Divide the numerator and denominator of the left and right side of the inequality by $x^2$ and then take limits. This limit will be $a$ which implies $a=3$. Left side of the inequality will go to $a$ because $5x^2\rightarrow 0$ as $x \rightarrow 0$ .

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Thank you Abhra for your answer! –  Lucas Jan 29 '13 at 12:33