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

Question

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.

Answer 1

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 $$\lim_{x \to 3^+} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}}=0$$

Answer 2

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 $$ \lim_{x\to 3}\frac{f(x)}{g(x)} = \lim_{x\to 3}\left(1 + \frac{1}{2\sqrt{x+1}}\right)\frac{\sqrt{x^2-9}}{x} = 0. $$

Answer 3

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 $$\frac{h-2+\sqrt{4+h}}{\sqrt{h(6+h)}}\sim_0 \frac{\cfrac{5 h}4}{\sqrt{6h}}=\frac {5\sqrt h}{4\sqrt6}\to 0.$$

Answer 4

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

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

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

Answer 5

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.

Answer 6

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 $$ \lim_{x \to 3} \frac{x-5+\sqrt{x+1}}{\sqrt{x^2-9}} = 0. $$

Answer 7

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

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