Thread: Calculate the limit (Question 1)

1. Calculate the limit (Question 1)

Hi,

Please tell me if my working for the following question is correct. Also, more importantly, I have some gaps in my math foundation, so please tell me what math topics I should revise/relearn for the questions I wrote in red. Thanks.

$\text{Calculate }\lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2+9)}$

$= \lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2-9) } \cdot \frac{5+ \sqrt(x^2+9)}{5+ \sqrt(x^2+9)}$ (I don't understand why + is used. Please tell me what topic I need to revise to fully understand this part)

$= \left (5- \sqrt(x^2+9) \right ) \cdot \left ( 5+ \sqrt(x^2+9) \right ) = 25-x^2 +9$ (Why do I need to multiply this?)

$= 16-x^2$

$= \left (4-x \right ) \left (4+x \right )$

$= \lim_{x \rightarrow 4} \frac{\left (4-x \right ) \left (5+ \sqrt (x^2+9) \right)}{\left (4-x \right ) \left (4+x \right )}$

$= \lim_{x \rightarrow 4 } \frac{5+ \sqrt(x^2+9)}{4+x}$

$= \frac{5+ \sqrt(4^2+9)}{4+4}$

$= \frac{5+ \sqrt(25)}{8}$

$\frac{10}{8}$

Edit: The question should be $\text{Calculate }\lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2+9)}$ I edited the original question above. Thanks TutorMeNate

2. Re: Calculate the limit (Question 1)

+ is used in order to remove the square root, refer to the identity (a+b)(a-b) = a^2-b^2

3. Re: Calculate the limit (Question 1)

Do you understand the following:
$\frac{1}{a+\sqrt{b}}=\frac{1}{a+\sqrt{b}}\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{a-\sqrt{b}}{a^2-b}~?$

4. Re: Calculate the limit (Question 1)

Originally Posted by piscoau
+ is used in order to remove the square root, refer to the identity (a+b)(a-b) = a^2-b^2
$= \lim_{x \rightarrow 4 } \frac{4-x}{5+ \sqrt(x^2+9) }$

I would change it to a minus instead, like this:

$= \lim_{x \rightarrow 4 } \frac{4-x}{5+ \sqrt(x^2+9) } \cdot \frac{5- \sqrt(x^2-9)}{5- \sqrt(x^2-9)}$

Is this correct? I will do some more reading on identity (a+b)(a-b) = a^2-b^2, thanks.

Originally Posted by Plato
Do you understand the following:
$\frac{1}{a+\sqrt{b}}=\frac{1}{a+\sqrt{b}}\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{a-\sqrt{b}}{a^2-b}~?$
Actually, yes I do.

5. Re: Calculate the limit (Question 1)

Wait a minute, I made a mistake in line 3:

$= \left (5- \sqrt(x^2-9) \right ) \cdot \left ( 5+ \sqrt(x^2+9) \right ) = 25-x^2 +9$

it is supposed to be

$= \left (5- \sqrt(x^2-9) \right ) \cdot \left ( 5+ \sqrt(x^2+9) \right ) = 25-x^2 +9$

6. Re: Calculate the limit (Question 1)

Originally Posted by sparky
Hi,

Please tell me if my working for the following question is correct. Also, more importantly, I have some gaps in my math foundation, so please tell me what math topics I should revise/relearn for the questions I wrote in red. Thanks.

$\text{Calculate }\lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2-9)}$

$= \lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2-9) } \cdot \frac{5+ \sqrt(x^2+9)}{5+ \sqrt(x^2+9)}$ (I don't understand why + is used. Please tell me what topic I need to revise to fully understand this part)
Well, its incorrect! Though perhaps you just misquoted it. What you should be doing is multiplying both numerator and denominator by $5+ \sqrt{x^2- 9}$, not $5+ \sqrt{x^2+ 9}$. That is, you do NOT change the sign inside the root.

$= \left (5- \sqrt(x^2+9) \right ) \cdot \left ( 5+ \sqrt(x^2+9) \right ) = 25-x^2 +9$ (Why do I need to multiply this?)
Okay, now you have " $x^2+ 9$" inside both roots where before you had " $x^2- 9$" in one of them.

$= 16-x^2$

$= \left (4-x \right ) \left (4+x \right )$

$= \lim_{x \rightarrow 4} \frac{\left (4-x \right ) \left (5+ \sqrt (x^2+9) \right)}{\left (4-x \right ) \left (4+x \right )}$

$= \lim_{x \rightarrow 4 } \frac{5+ \sqrt(x^2+9)}{4+x}$

$= \frac{5+ \sqrt(4^2+9)}{4+4}$

$= \frac{5+ \sqrt(25)}{8}$

$\frac{10}{8}$

7. Re: Calculate the limit (Question 1)

Originally Posted by sparky
Wait a minute, I made a mistake in line 3:

$= \left (5- \sqrt(x^2-9) \right ) \cdot \left ( 5+ \sqrt(x^2+9) \right ) = 25-x^2 +9$

it is supposed to be

$= \left (5- \sqrt(x^2-9) \right ) \cdot \left ( 5+ \sqrt(x^2+9) \right ) = 25-x^2 +9$
No, that's still wrong. Ether $(5-\sqrt{x^2- 9})(5+ \sqrt{x^2- 9})= 25- (x^2- 9)= 34- x^2$
or $(5-\sqrt{x^2+ 9})(5+ \sqrt{x^2+9})= 25- (x^2+ 9)= 16- x^2$

But you can't have different signs inside the square roots.

8. Re: Calculate the limit (Question 1)

Originally Posted by HallsofIvy
Well, its incorrect! Though perhaps you just misquoted it. What you should be doing is multiplying both numerator and denominator by $5+ \sqrt{x^2- 9}$, not $5+ \sqrt{x^2+ 9}$. That is, you do NOT change the sign inside the root.
Hey HallsofIvy thanks. So it should be:

$= \left (5 - \sqrt(x^2-9) \right ) \cdot \left ((5+ \sqrt(x^2-9) \right )$

$= 25 - x^2 - 9$

$= 16 - x^2$

Now my question is why do I have to change the negative sign on the outside?

9. Re: Calculate the limit (Question 1)

Hi Sparky,

Two things:

First, you change the sign because you are multiplying by the conjugate, which answers your initial question of what you should look up and review.

Second, and to be very blunt, you did a lot of pointless work. You're trying to find the limit as x approaches 4. Before you ever try to simplify a limit, always try to plug it in first. If it is undefined, then proceed to simplify. In this case, you get 0 when you plug in 4, meaning the answer is 0. You can also check this by using a graphing calculator.

I hope this helps!

-Nathan

10. Re: Calculate the limit (Question 1)

I reworked the problem (please see below) and got a different answer $\left (\frac{5+ \sqrt(7)}{8 } \right )$ than from what we got in class $\left (\frac{10}{8} \right )$. What is the correct answer? What did I do wrong? How can I tell whats the correct answer?

$\text{Calculate }\lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2-9)}$

$= \lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2-9) } \cdot \frac{5+ \sqrt(x^2-9)}{5+ \sqrt(x^2-9)}$

$= \left (5- \sqrt(x^2-9) \right ) \cdot \left ( 5+ \sqrt(x^2-9) \right )$

$= 25-x^2 +9$

$= 16-x^2$

$= \left (4-x \right ) \left (4+x \right )$

$= \lim_{x \rightarrow 4} \frac{\left (4-x \right ) \left (5+ \sqrt (x^2-9) \right)}{\left (4-x \right ) \left (4+x \right )}$

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

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

$= \frac{5+ \sqrt(7)}{8 }$ in class we got $\frac{10}{8}$

11. Re: Calculate the limit (Question 1)

Considering you got 10/8 in class, you copied the problem wrong. It should be sqrt(x^(2)+9) in the original problem. If you solve it with that change, then you should get 10/8.

I hope this helps!

-Nathan

12. Re: Calculate the limit (Question 1)

Here is a LaTeX TIP
[TEX]5+\sqrt{x^2+9}[/TEX] gives $5+\sqrt{x^2+9}$.

13. Re: Calculate the limit (Question 1)

Originally Posted by Plato
Here is a LaTeX TIP
[TEX]5+\sqrt{x^2+9}[/TEX] gives $5+\sqrt{x^2+9}$.

14. Re: Calculate the limit (Question 1)

Originally Posted by TutorMeNate
Considering you got 10/8 in class, you copied the problem wrong. It should be sqrt(x^(2)+9) in the original problem. If you solve it with that change, then you should get 10/8.

I hope this helps!

-Nathan
Jezzz, you were right! Thanks. Here is my corrected version:

$\text{Calculate }\lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt{x^2+9}}$

$= \lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt{x^2+9}} \cdot \frac{5+ \sqrt{x^2+9}}{5+ \sqrt{x^2+9}}$

$= \left (5- \sqrt{x^2+9} \right ) \cdot \left ( 5+ \sqrt{x^2+9} \right )$

$= 25-x^2 +9$

$= 16-x^2$

$= \left (4-x \right ) \left (4+x \right )$

$= \lim_{x \rightarrow 4} \frac{\left (4-x \right ) \left (5+ \sqrt {x^2+9} \right)}{\left (4-x \right ) \left (4+x \right )}$

$= \lim_{x \rightarrow 4 } \frac{5+ \sqrt{x^2+9}}{4+x}$

$= \frac{5+ \sqrt{4^2+9}}{4+4}$

$= \frac{5+ \sqrt{25}}{8}$

$= \frac{10}{8}$