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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?

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  • $\begingroup$ Apply the squeeze theorem. $\endgroup$
    – Mark Viola
    Commented Feb 16, 2017 at 16:09
  • $\begingroup$ Interesting, I hadn't considered that but I only have one side of the interval. $\endgroup$ Commented Feb 16, 2017 at 16:14
  • $\begingroup$ How about $|f(x) - \sqrt{x^2 + 9}| \leq x^2 + |3 - \sqrt{x^2 + 9}|$? $\endgroup$
    – Hopeless
    Commented Feb 16, 2017 at 16:14

3 Answers 3

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"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!

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  • $\begingroup$ This still doesn't solve the limit $\endgroup$ Commented Feb 16, 2017 at 16:18
  • $\begingroup$ Is there any reason as to why the first statement is defined? $\endgroup$ Commented Feb 16, 2017 at 16:41
  • $\begingroup$ Gary, not that $|x|\le y$ implies that $-y\le x\le y$. $\endgroup$
    – Mark Viola
    Commented Feb 16, 2017 at 16:43
  • $\begingroup$ @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 $\endgroup$
    – Mark Viola
    Commented Feb 16, 2017 at 16:51
  • $\begingroup$ May I also ask as to how the simplification within the braces in step 2 was achieved ? $\endgroup$ Commented Feb 16, 2017 at 16:52
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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.

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  • $\begingroup$ How did you get the above expression? $\endgroup$ Commented Feb 16, 2017 at 16:22
  • $\begingroup$ @GaryAndrews30 $|f(x)-\sqrt{x^2 +9}| = |f(x)-3 + 3 - \sqrt{x^2 + 9}| \le |f(x)-3| + |3 - \sqrt{x^2+9}|$ $\endgroup$
    – user384138
    Commented Feb 16, 2017 at 16:23
  • $\begingroup$ 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? $\endgroup$ Commented Feb 16, 2017 at 16:33
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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 $\quad \lim_{x \rightarrow 0}L = 0$

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