# How do you evaluate $\lim_{x \to 3^+} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}$? [closed]

Evaluate $$\lim_{x \to 3^+} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}$$

I tried substitution first. (It won't work for this one.) When finding a limit of a fraction and in doubt, rationalize either the numerator or denominator.

• Is there a question? Commented Sep 17, 2018 at 21:28
• how to calculate Limit[(x - 5 + Sqrt[1 + x])/Sqrt[x^2 - 9], x -> 3] Commented Sep 17, 2018 at 21:30
• Please read this tutorial on how to typeset mathematics on this site, then edit your question to show what you have attempted and to explain where you are stuck. Commented Sep 17, 2018 at 21:32
• rationalize the numerator multiplying by $\frac {x-5 - \sqrt{x+1}}{x-5 - \sqrt{x+1}}$ we should expect to get an $(x-3)$ factor in the numerator. This will partially cancel with what is in the denominator giving $\sqrt{x-3}(g(x))$ where $g(x)$ is continuous at 3. Commented Sep 17, 2018 at 21:38
• Strictly speaking, you probably should have $\lim_{x\to 3^+}$ ( because if $-3<x <3$, we might be troubled by how to evaluate $\sqrt{x^2-9}$...) Commented Sep 17, 2018 at 21:45

The limit does not exist because as you notice we can only find the right limit because the left limit does not make sense at $x=3$.

$$\lim_{x \to 3^+} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}$$

could be found by multiplying top and bottom by $x-5-\sqrt {x+1}$ where the top factors as $(x-3)(x+8)$ and after elimination of $\sqrt {x-3}$ from the top and bottom we come up with $$\lim_{x \to 3^+} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}=0$$

Here is how to rationalize the numerator: \begin{align} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}\frac{x-5-\sqrt{x+1}}{x-5-\sqrt{x+1}} &=\frac{x^2-11x+24}{\sqrt{x-3}\sqrt{x+3}\left(x-5-\sqrt{x+1}\right)}\\ &=\frac{x-3}{\sqrt{x-3}}\frac{x-8}{\sqrt{x+3}\left(x-5-\sqrt{x+1}\right)} \end{align} Now it is easier to take the limit.

Set $x=3+h$, $\;h\to 0$. The expression becomes $$\frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}=\frac{h-2+\sqrt{4+h}}{\sqrt{h(6+h)}}=$$ Now $\;\sqrt{4+h}=2\sqrt{1+\frac h4}=2\bigl(1+\frac h8+o(h)\bigr)$, so the numerator is $$h-2+2+\frac h4+o(h)=\frac{5h}4+o(h)\sim_0\frac{5h}4.$$ On the other hand $h(6+h)\sim_06h$, so $$\frac{h-2+\sqrt{4+h}}{\sqrt{h(6+h)}}\sim_0 \frac{\cfrac{5 h}4}{\sqrt{6h}}=\frac {5\sqrt h}{4\sqrt6}\to 0.$$

• I had the same idea!
– user
Commented Sep 17, 2018 at 21:48
• @gimusi: Finally, I thought finding and using equivalents was simpler. That being said, great minds think together ;o) Commented Sep 17, 2018 at 21:50
• I've also modified a little bit without change of variable! But the idea is the same.
– user
Commented Sep 17, 2018 at 21:57

Well. Let consider the following: \begin{align} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}} &= \frac{x-3-2+\sqrt{x+1}}{\sqrt{x-3}\sqrt{x+3}} = \frac{x-3}{\sqrt{x-3}\sqrt{x+3}} + \frac{-2+\sqrt{x+1}}{\sqrt{x-3}\sqrt{x+3}} \\ &= \frac{\sqrt{x-3}}{\sqrt{x+3}} + \frac{\frac{-4+x+1}{2+\sqrt{x+1}}}{\sqrt{x-3}\sqrt{x+3}} \\ &= \frac{\sqrt{x-3}}{\sqrt{x+3}} + \frac{\sqrt{x-3}}{\sqrt{x+3}}\frac{1}{2+\sqrt{x+1}}\\ &= \frac{\sqrt{x-3}}{\sqrt{x+3}}\left(1 + \frac{1}{2+\sqrt{x+1}}\right). \end{align} Therefore, by letting $x \to 3$, we have $$\lim_{x \to 3} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}} = 0.$$

By binomial expansion we have

$$\sqrt{x+1}=\sqrt{4+(x-3)}=2\sqrt{1+(x-3)/4}=2+(x-3)/4+o(x-3)$$

and therefore

$$\frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}=\frac{x-5+2+(x-3)/4+o(x-3)}{x^2-9}\sqrt{x^2-9}=$$

$$=\frac{5(x-3)/4+o(x-3)}{x^2-9}\sqrt{x^2-9}=\frac{5/4+o(1)}{x+3}\sqrt{x^2-9}\to 0$$

$\frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}} = \frac {x-5+\sqrt{x+1}}{\sqrt{(x-3)(x+3)}}$

If we can factor out $\sqrt{x -3}$ in the numerator we are good.

$x - 5 + \sqrt {x+1} = x -3 + (-2 + \sqrt{x+1})$ so

$\frac {x-5+\sqrt{x+1}}{\sqrt{(x-3)(x+3)}}= \frac {x-3}{\sqrt{x -3}\sqrt{x+3}} + \frac {-2 + \sqrt{x+1}}{\sqrt{x -3}\sqrt{x+3}} = \frac {\sqrt {x-3}}{\sqrt{x + 3}} + \frac {-2 + \sqrt{x+1}}{\sqrt{x -3}\sqrt{x+3}}$

Now we know the first term will go to $0$ and we can rationalize the numerator of the second term via $(-2 + \sqrt{x+1})(-2 - \sqrt{x+1}) = 4 - (x+1) = 3-x$.

So we have:

$\frac {x-5+\sqrt{x+1}}{\sqrt{(x-3)(x+3)}}= \frac {\sqrt {x-3}}{\sqrt{x + 3}} - \frac {x-3}{\sqrt{x -3}\sqrt{x+3}(-2 - \sqrt{x+1})}= \frac {\sqrt {x-3}}{\sqrt{x + 3}} - \frac {\sqrt{x -3}}{\sqrt{x+3}(-2 - \sqrt{x+1})}$

And $\lim\limits_{x\to 3}\frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}=\lim\limits_{x\to 3}\frac {\sqrt {x-3}}{\sqrt{x + 3}} - \frac {\sqrt{x -3}}{\sqrt{x+3}(-2 - \sqrt{x+1})}= \frac 0{\sqrt 6} - \frac 0{-4*\sqrt 6} = 0$

You can use L'Hôpital's rule. It states that $$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$$ In your case $f(x) = x - 5 + \sqrt{x+1}$ and $g(x) =\sqrt{x^2-9}$. Hence $$\lim_{x\to 3}\frac{f(x)}{g(x)} = \lim_{x\to 3}\left(1 + \frac{1}{2\sqrt{x+1}}\right)\frac{\sqrt{x^2-9}}{x} = 0.$$

• A student at this level almost surely doesn't know L'Hospital's rule.
– user296602
Commented Sep 17, 2018 at 21:41
• @N.F.Taussig Fixed it, thank you! Commented Sep 17, 2018 at 21:44