# Evaluate the given limit

Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find

$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$

Can the function $f(x)$ be considered as $x^2 + 3$ and go about evaluating the limit using the Limit laws?

• Apply the squeeze theorem. – Mark Viola Feb 16 '17 at 16:09
• Interesting, I hadn't considered that but I only have one side of the interval. – Gary Andrews30 Feb 16 '17 at 16:14
• How about $|f(x) - \sqrt{x^2 + 9}| \leq x^2 + |3 - \sqrt{x^2 + 9}|$? – Hopeless Feb 16 '17 at 16:14

"Can the function $f(x)$ be considered as $x^2 + 3$ and go about solving the limit using the Limit laws?"

No, since we have only that $|f(x)-3|\le x^2\implies 3-x^2\le f(x)\le 3+x^2$.

But we can proceed by using $\color{blue}{f(x)-3=O(x^2)}$, where we are using the ("Big O notation").

Then, we can evaluate the limit of interest by writing

\begin{align} \frac{f(x)-\sqrt{x^2+9}}{x}&=\frac{f(x)-3\left(1+\frac{x^2}{9}\right)^{1/2}}{x}\\\\ &=\frac{f(x)-3\left(1+\color{red}{\frac12 \frac{x^2}{9}+O(x^4)}\right)}{x}\\\\ &=\frac{\color{blue}{\left(f(x)-3\right)}+\color{red}{O(x^2)}}{x}\\\\ &=\frac{\color{blue}{O(x^2)}+\color{red}{O(x^2)}}{x}\\\\ &=O(x)\to 0\,\,\text{as}\,\,x\to 0 \end{align}

And we are done!

• This still doesn't solve the limit – Simply Beautiful Art Feb 16 '17 at 16:18
• Is there any reason as to why the first statement is defined? – Gary Andrews30 Feb 16 '17 at 16:41
• Gary, not that $|x|\le y$ implies that $-y\le x\le y$. – Mark Viola Feb 16 '17 at 16:43
• @SimplyBeautifulArt It's a bit embarrassing. I had edited the question and not digested it fully. I've edited accordingly. Thank you for alerting me! -Mark – Mark Viola Feb 16 '17 at 16:51
• May I also ask as to how the simplification within the braces in step 2 was achieved ? – Gary Andrews30 Feb 16 '17 at 16:52

We have $|f(x)-\sqrt{x^2 +9}| = |f(x)-3 + 3 - \sqrt{x^2 + 9}| \le |f(x)-3| + |3 - \sqrt{x^2+9}| \le x^2 + |3 - \sqrt{x^2 + 9}|$. So:

$$\left|\frac{f(x)-\sqrt{x^2+9}}{x}\right| \le |x| + \left| \frac{\sqrt{x^2+9}-3}{x}\right|$$

Now squeeze.

• How did you get the above expression? – Gary Andrews30 Feb 16 '17 at 16:22
• @GaryAndrews30 $|f(x)-\sqrt{x^2 +9}| = |f(x)-3 + 3 - \sqrt{x^2 + 9}| \le |f(x)-3| + |3 - \sqrt{x^2+9}|$ – user384138 Feb 16 '17 at 16:23
• The given interval should be simplified to f(x) to squeeze, the left side of the interval would be 0. How can the expression be simplified? – Gary Andrews30 Feb 16 '17 at 16:33

Apply law |a+b| <= |a| + |b|

Let $L = \frac{f(x)-\sqrt{x^2+9}}{x}$

$$|L| = \lvert\frac{f(x) - 3 + 3 - \sqrt{x^2+9}}{x}\rvert \le |\frac{f(x)-3}{x}| + |\frac{3 - \sqrt{x^2+9}}{x}| \\ \le |\frac{x^2}{x}| + |\frac{(3-\sqrt{x^2+9})(3+\sqrt{x^2+9})}{x(3+\sqrt{x^2+9})}| = |x| + |\frac{x}{3+\sqrt{x^2+9}}|$$

Hence, $$\lim_{x \rightarrow 0}{|L|} \le \lim_{x \rightarrow 0}{(|x| + |\frac{x}{3+\sqrt{x^2+9}}|)} = 0 \quad (1)$$

Since $|L| \ge 0 \quad \forall x$, we also have $\lim_{x \rightarrow 0}{|L|} \ge 0 \quad (2)$

From (1) and (2) we have $\lim_{x \rightarrow 0}{|L|} = 0$, or $\quad \lim_{x \rightarrow 0}L = 0$