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Math Help - Calculate the limit (Question 1)

  1. #1
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    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
    Last edited by sparky; September 14th 2011 at 05:00 PM.
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  2. #2
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    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
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  3. #3
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    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}~?
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  4. #4
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    Re: Calculate the limit (Question 1)

    Quote Originally Posted by piscoau View Post
    + is used in order to remove the square root, refer to the identity (a+b)(a-b) = a^2-b^2
    Ok, so if I had this instead:
    = \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.

    Quote Originally Posted by Plato View Post
    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.
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  5. #5
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    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
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  6. #6
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    Re: Calculate the limit (Question 1)

    Quote Originally Posted by sparky View Post
    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}
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  7. #7
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    Re: Calculate the limit (Question 1)

    Quote Originally Posted by sparky View Post
    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.
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  8. #8
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    Re: Calculate the limit (Question 1)

    Quote Originally Posted by HallsofIvy View Post
    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?
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  9. #9
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    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
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  10. #10
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    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}
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  11. #11
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    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
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  12. #12
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    Re: Calculate the limit (Question 1)

    Here is a LaTeX TIP
    [TEX]5+\sqrt{x^2+9}[/TEX] gives 5+\sqrt{x^2+9}.
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  13. #13
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    Re: Calculate the limit (Question 1)

    Quote Originally Posted by Plato View Post
    Here is a LaTeX TIP
    [TEX]5+\sqrt{x^2+9}[/TEX] gives 5+\sqrt{x^2+9}.
    Yeah, that was helpful thanks!
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  14. #14
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    Re: Calculate the limit (Question 1)

    Quote Originally Posted by TutorMeNate View Post
    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}
    Last edited by sparky; September 14th 2011 at 04:33 PM.
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