# 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. Commented Feb 16, 2017 at 16:09
• Interesting, I hadn't considered that but I only have one side of the interval. Commented Feb 16, 2017 at 16:14
• How about $|f(x) - \sqrt{x^2 + 9}| \leq x^2 + |3 - \sqrt{x^2 + 9}|$? Commented Feb 16, 2017 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 Commented Feb 16, 2017 at 16:18
• Is there any reason as to why the first statement is defined? Commented Feb 16, 2017 at 16:41
• Gary, not that $|x|\le y$ implies that $-y\le x\le y$. Commented Feb 16, 2017 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 Commented Feb 16, 2017 at 16:51
• May I also ask as to how the simplification within the braces in step 2 was achieved ? Commented Feb 16, 2017 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? Commented Feb 16, 2017 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
Commented Feb 16, 2017 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? Commented Feb 16, 2017 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 $\quad \lim_{x \rightarrow 0}L = 0$