3
$\begingroup$

As stated in the title, I need to prove that $\lim_{x\to 2} \sqrt{x^2+5} = 3$ using only the precise definition of a limit.


For any given $\varepsilon \gt 0$, there exists a $\delta = $

Such that $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} \rvert \lt \varepsilon $

I've attempted to convert $\sqrt{x^2+5}$ into the following:

$\sqrt{x^2+5} = \sqrt{(x^2-4)+9}\\ \sqrt{x^2+5} = \sqrt{(x^2-4x+4)+4x +1} = \sqrt{(x-2)^2+4(x-2)+9} $

I have a hunch that I am heading in the wrong direction. Can I get some advice on what I might be doing wrong?

$\endgroup$

3 Answers 3

4
$\begingroup$

You are on the right track.

Also you can do this:

Let $\epsilon >0$

If $\delta<1$ then $1<x<3$ thus $|x+2|<5$

\begin{align}|\sqrt{x^2+5}-3|&=\frac{(\sqrt{x^2+5}-3)(\sqrt{x^2+5}+3)}{\sqrt{x^2+5}+3}\\&=\frac{|x^2-4|}{\sqrt{x^2+5}+3} \\&\leq \frac{|x+2||x-2|}{3}\\&\leq \frac{5}{3}|x-2|\end{align}

Now take $\delta \leq \min\{\frac{3}{5} \epsilon,1\}$ and you are done.

$\endgroup$
2
$\begingroup$

You're doing fine! So, in order to have$$\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|<\varepsilon,$$all you need is to have\begin{multline}\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|<\\<\varepsilon.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|,\end{multline}or, in other words,$$\bigl|(x-2)^2+4(x-2)\bigr|<\varepsilon.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|.$$In order for this to happen, all you need is that $\bigl|(x-2)^2+4(x-2)\bigr|<3\varepsilon$. So, pick $\delta$ such that $\delta\leqslant\frac38\varepsilon$ and also $\delta\leqslant\sqrt{\frac32\varepsilon}$. Then, if $|x-2|<\delta$,$$\bigl|(x-2)^2+4(x-2)\bigr|\leqslant|x-2|^2+4|x-2|\leqslant\frac32\varepsilon+\frac32\varepsilon=3\varepsilon.$$

$\endgroup$
2
$\begingroup$

We should prove that,for any given $ε>0$, there exists a $δ$ if $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} -3\rvert \lt \varepsilon$ $$\left| \sqrt { { x }^{ 2 }+5 } -3 \right| =\left| \frac { { x }^{ 2 }-4 }{ \sqrt { { x }^{ 2 }+5 } +3 } \right| =\left| \frac { { \left( x-2 \right) }\left( x-2+4 \right) }{ \sqrt { { x }^{ 2 }+5 } +3 } \right| <\frac { \left| { \left( x-2 \right) }^{ 2 }+4\left( x-2 \right) \right| }{ 3 } <\frac { { \left| x-2 \right| }^{ 2 }+4\left| x-2 \right| }{ 3 } <\frac { { \delta }^{ 2 }+4\delta }{ 3 } =\epsilon $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .