# Proof verification: $\lim_{x\to 2} \frac{\sqrt{x^2 + 5} – 3}{x – 2} = \frac23$

analysiscalculusproof-writingsolution-verification

The question is as follows:

Prove that $$\displaystyle\lim_{x\to 2} \dfrac{\sqrt{x^2 + 5} – 3}{x – 2} = \dfrac23$$.

My proof is:

Fix $$\varepsilon > 0$$.

Note that $$\begin{array}{rcl}\left|\dfrac{\sqrt{x^2 + 5} – 3}{x – 2} – \dfrac23\right| &=& \left|\dfrac{x + 2}{\sqrt{x^2 + 5} + 3} – \dfrac23\right|\\&=& \dfrac13\left|\dfrac{3x – 2\sqrt{x^2 + 5}}{3 + \sqrt{x^2 + 5}}\right|\\&=& \dfrac13 \left|\dfrac{5x^2 – 20}{\left(3 + \sqrt{x^2 + 5}\right)\left(3x + 2\sqrt{x^2 + 5}\right)}\right|\\&=& \dfrac{5|x – 2|\cdot|x + 2|}{3\left|3 + \sqrt{x^2 + 5}\right|\cdot \left|3x + 2\sqrt{x^2 + 5}\right|}\end{array}$$

Pick $$\delta = \min\left\{1, \dfrac{21\varepsilon}5\right\}$$. Suppose that $$0 < |x – 2| < \delta < 1$$. Then $$1 < x < 3$$. Therefore, $$|x + 2| < 5$$, $$\left|3 + \sqrt{x^2 + 5}\right| > 3 + \sqrt 6 > 5$$, and $$\left|3x + 2 \sqrt{x^2 + 5}\right| > 3 + 2\sqrt 6 > 7$$.

Therefore,
$$\left|\dfrac{\sqrt{x^2 + 5} – 3}{x – 2} – \dfrac23\right| < \dfrac{5 \cdot 5}{3 \cdot 5 \cdot 7} |x – 2| < \dfrac5{21} \delta < \varepsilon$$

Hence, $$\displaystyle\lim_{x\to 2} \dfrac{\sqrt{x^2 + 5} – 3}{x – 2} =\dfrac23$$.

Is my proof correct?

Yes, you have the main idea correct.

Though, there seems to be too much formalism without explaining in words, which may obscure your proof.

One key step you make is to rewrite the function as $$\frac{\sqrt{x^2+5}-3}{x-2}=\frac{x^2+5-9}{(x-2)(\sqrt{x^2+5}+3)} =\frac{x+2}{\sqrt{x^2+5}+3}\;.$$ At this point, you can simply apply one of the limit laws and continuity of the functions to conclude that $$\lim_{x\to 2}\frac{x+2}{\sqrt{x^2+5}+3}=\frac{4}{6}=\frac23\;.$$

Students usually confuse "formal/rigorous proof" with proofs written in terms of $$\epsilon$$-$$\delta$$.

If one does want to proceed with an $$\epsilon$$-$$\delta$$ proof, then one needs to estimate the quantity $$\left|\frac{x+2}{\sqrt{x^2+5}+3}-\frac23\right|\;\tag{1}$$

Instead of handling this on an ad hoc way, one can follow the idea of proving the quotient law for limits to estimate (1) as \begin{align} \left|\frac{x+2}{\sqrt{x^2+5}+3}-\frac46\right| =& \frac{|6f(x)-4g(x)|}{|6g(x)|} \\ =& \frac{|6f(x)-6\cdot 4+6\cdot 4- 4g(x)|}{|6g(x)|}\\ \le&\frac{6|f(x)-4|+4|g(x)-6|}{|6g(x)|}\tag{2} \end{align} where $$f(x)=x+2$$ and $$g(x)=\sqrt{x^2+5}+3$$. It is then clear that one should use the following facts to get a desired $$\delta$$:

• $$f$$ and $$g$$ are continuous at $$x=2$$;
• $$g$$ is bounded away from $$0$$ near $$x=2$$.

To spell out the formalism, one can take $$\delta=\min(\delta_1,\delta_2,\delta_3)$$ such that \begin{align} |f(x)-4|<\epsilon\qquad\textrm{whenever } |x-2|<\delta_1\\ |g(x)-6|<\epsilon\qquad\textrm{whenever } |x-2|<\delta_2\\ |g(x)|\ge |g(x)-6+6|\ge 6-|g(x)-6|\ge 1\qquad\textrm{whenever } |x-2|<\delta_3\tag{3} \end{align} Combining with (2) one has $$\left|\frac{x+2}{\sqrt{x^2+5}+3}-\frac46\right| \le \frac{10\epsilon}{6}\;.$$