# Thread: Calculate the limit (Question 2)

1. ## Calculate the limit (Question 2)

I'm stuck on the last line and I'm not sure how to finish:

$\text{Calculate }\lim_{x \rightarrow-2 } \frac{x+2}{\sqrt(x^+5) -3}$

$= \lim_{x \rightarrow-2 } \frac{x+2}{\sqrt(x^+5) -3 } \cdot \frac{\sqrt (x^2-5) + 3}{\sqrt (x^2-5) + 3}$

$= \left (\sqrt (x^2 +5) -3 \right ) \cdot \left (\sqrt (x^2 -5) +3 \right )$

$= x^2 + 5 - 9$

$= x^2 - 4$

$= (x+2)(x-2)$

$= \lim_{x \rightarrow-2 } \frac{(x+2) \cdot \sqrt (x^2 -5) + 3}{(x+2)(x-2)}$

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

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

$= \frac{\sqrt (-1)+3}{-4 }$Stuck here, adding square root of -1+3

2. ## Re: Calculate the limit (Question 2)

Originally Posted by sparky
I'm stuck on the last line and I'm not sure how to finish:

$\text{Calculate }\lim_{x \rightarrow-2 } \frac{x+2}{\sqrt(x^+5) -3}$
For this you just need to substitute -2 for x to get the limit is 0, unless everything in the denominator is under the square root

CB

3. ## Re: Calculate the limit (Question 2)

$\text{Calculate } \lim_{x\rightarrow-2} \frac{x+2}{\sqrt(x^2 + 5)-3}$

$= \frac{-2+2}{\sqrt(4^2 + 5)-3}$

$=\frac{0}{3-3}$

$= 0$

CaptainBlack, I have three questions:

1. What does 0 really mean?
2. What is the importance of plugging in -2 first?
3. What if the answer was not 0? than what do I do?

4. ## Re: Calculate the limit (Question 2)

The importance of plugging in -2 first is that you can determine if you're dealing with an undetermined form or not, in this case you see you get the undetermined form: $\frac{0}{0}$.

Calculating:
$\lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(\sqrt{x^2+5}-3)\cdot (\sqrt{x^2+5}+3)}$
$=\lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{x^2-4}$
$=\lim_{x\to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(x-2)\cdot (x+2)}$
$=\lim_{x\to -2} \frac{\sqrt{x^2+5}+3}{x-2}$

Now plug in -2 again, so what's the limit?

5. ## Re: Calculate the limit (Question 2)

Originally Posted by Siron
The importance of plugging in -2 first is that you can determine if you're dealing with an undetermined form or not, in this case you see you get the undetermined form: $\frac{0}{0}$.

Calculating:
$\lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(\sqrt{x^2+5}-3)\cdot (\sqrt{x^2+5}+3)}$
$=\lim_{x \to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{x^2-4}$
$=\lim_{x\to -2} \frac{(x+2)\cdot (\sqrt{x^2+5}+3)}{(x-2)\cdot (x+2)}$
$=\lim_{x\to -2} \frac{\sqrt{x^2+5}+3}{x-2}$

Now plug in -2 again, so what's the limit?
Thanks Siron for pointing out where I went wrong in my working. Here is what I got:

$=\frac{\sqrt {-2^2 + 5} + 3}{-2-2}$

$=\frac{\sqrt {9} + 3}{-4}$

$=\frac{6}{-4}$

$= - \frac{3}{2}$

I want to be clear on this and correct me if I am wrong: if I get an undetermined value when I plug in the value (e.g. -2), then I have to calculate the limit the long way, which is what we just did. Otherwise, if I don't get an undertermined value, then the answer is the limit?

6. ## Re: Calculate the limit (Question 2)

Originally Posted by sparky
$\text{Calculate } \lim_{x\rightarrow-2} \frac{x+2}{\sqrt(x^2 + 5)-3}$

$= \frac{-2+2}{\sqrt(4^2 + 5)-3}$

$=\frac{0}{3-3}$

$= 0$

CaptainBlack, I have three questions:

1. What does 0 really mean?
2. What is the importance of plugging in -2 first?
3. What if the answer was not 0? than what do I do?
Where did the x^2 suddenly appear from? If what you now post is what you really want then plugging -2 in for x won't work because what you get is indeterminate.

You will either have to use L'Hopital's rule or the substitution u=x+2 and consider the resulting limit as u goes to zero

CB

7. ## Re: Calculate the limit (Question 2)

Sorry Captain Black, I made a mistake with the question. The correct question is:

$\text{Calculate } \lim_{x \rightarrow -2} \frac{x+2}{ \sqrt{x^2 + 5}-3}$

$\lim_{x \rightarrow -2} \frac{x+2}{ \sqrt{x^2 + 5}-3 } \cdot \frac{\sqrt{x^2 + 5}+3}{\sqrt{x^2 + 5}+3}$

$(\sqrt{x^2+5} -3) \cdot (\sqrt{x^2+5}+3)$

$x^2 +5 -9$

$x^2 - 4$

$(x+2)(x-2)$

$\lim_{x \rightarrow -2 } \frac{(x+2) \cdot (\sqrt{x^2 + 5}+3)}{(x+2) \cdot (x-2)}$

$\lim_{x \rightarrow -2 } \frac{\sqrt{x^2+5}+3}{x-2}$

$= \frac{\sqrt{-2^2+5}+3}{-2-2}$

$\frac{6}{-4}$

$- \frac{3}{2}$

How is my working and answer?

8. ## Re: Calculate the limit (Question 2)

This working and answer is fine!
Note: make sure you use brackets where necessary, because $(-2)^2=4$ but $-2^2=-4$.