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Math Help - Quick help in Limits

  1. #1
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    Quick help in Limits

    Can anyone please help explain how to go about evaluating a few limits. Steps and procedures in solving them is much appreciated:

     lim (h goes to 0) 5^h-2^h/h

     lim (ygoes to -3) y^2-y-12/y+3

    again, help is much appreciated.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by cjmac87 View Post
    Can anyone please help explain how to go about evaluating a few limits. Steps and procedures in solving them is much appreciated:

     lim (h goes to 0) 5^h-2^h/h

     lim (ygoes to -3) y^2-y-12/y+3

    again, help is much appreciated.
    I guess it's this :

    \lim_{h \to 0} \frac{5^h-2^h}{h}

    \lim_{y \to -3} \frac{y^2-y-12}{y+3}

    (click on the pictures to get the code)

    For the second one, note that y^2-y-12=(y+3)(y-4)

    For the first one, are you familiar with L'Hospital's rule ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cjmac87 View Post
    Can anyone please help explain how to go about evaluating a few limits. Steps and procedures in solving them is much appreciated:

     lim (h goes to 0) 5^h-2^h/h
    i'm sure there's an easier way, and someone will point it out soon because everyone hates what i am about to suggest.

    try applying L'Hopital's rule

    *runs and jumps out the window to avoid the angry mob of mathematicians*

     lim (ygoes to -3) y^2-y-12/y+3

    again, help is much appreciated.
    Hint: note that this is the same as \lim_{y \to -3} \frac {(y + 3)(y - 4)}{(y + 3)}

    now what?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    For the first one, are you familiar with L'Hospital's rule ?
    haha, we stand together! i am safe now. the mob won't hurt you, you can protect me
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  5. #5
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    Thanks for the code.

    And no, I am not familiar with that rule at all...

    However, as you said to factor the one problem, i came up with an answer -7 by canceling two (y+3). According to my book, the answer is 0...
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  6. #6
    Moo
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    Quote Originally Posted by Jhevon View Post
    haha, we stand together! i am safe now. the mob won't hurt you, you can protect me
    You could've talked about the moob

    Quote Originally Posted by cjmac87 View Post
    Thanks for the code.

    And no, I am not familiar with that rule at all...

    However, as you said to factor the one problem, i came up with an answer -7 by canceling two (y+3). According to my book, the answer is 0...
    Hmmm well, I'm pretty sure the answer is -7... Did you look at the right answer ?


    For the first one, do you know this formula :
    \lim_{h \to a} \frac{f(a+h)-f(a)}{h}=f '(a)

    Write your limit this way :

    \lim_{h \to 0} \frac{5^h-2^h}{h} =\lim_{h \to 0} \frac{(5^h\overbrace{-1)+(1}^{\text{this is 0}}-2^h)}{h}=\lim_{h \to 0} \frac{5^h-1}{h}-\lim_{h \to 0} \frac{2^h-1}{h}=\lim_{h \to 0} \frac{5^h-5^0}{h}-\lim_{h \to 0} \frac{2^h-2^0}{h}

    Apply the formula I gave you with a=0 and try to find the function f for the two limits
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cjmac87 View Post
    Thanks for the code.

    And no, I am not familiar with that rule at all...
    L'Hopital's rule says

    if \lim_{x \to a} \frac {f(x)}{g(x)} \to \frac 00 \mbox{ or } \frac {\infty}{\infty}, then \lim_{x \to a} \frac {f(x)}{g(x)} = \lim_{x \to a} \frac {f'(x)}{g'(x)}

    However, as you said to factor the one problem, i came up with an answer -7 by canceling two (y+3). According to my book, the answer is 0...
    you are correct. the book is wrong, or you are looking in the wrong place
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    You could've talked about the moob
    oh yeah, if that's the case, we're both in trouble. just tell them you're the queen of MHF and so hurting you would cause an all out war. in addition, you have diplomatic immunity, so even the authorities have no business messing with you

    For the first one, do you know this formula :
    \lim_{h \to a} \frac{f(a+h)-f(a)}{h}=f '(a)

    Write your limit this way :

    \lim_{h \to 0} \frac{5^h-2^h}{h} =\lim_{h \to 0} \frac{(5^h\overbrace{-1)+(1}^{\text{this is 0}}-2^h)}{h}=\lim_{h \to 0} \frac{5^h-1}{h}-\lim_{h \to 0} \frac{2^h-1}{h}=\lim_{h \to 0} \frac{5^h-5^0}{h}-\lim_{h \to 0} \frac{2^h-2^0}{h}

    Apply the formula I gave you with a=0 and try to find the function f for the two limits
    i thought of a similar way after i posted
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  9. #9
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    Well, it's not that hard when actually having in mind a limit like \underset{x\to 0}{\mathop{\lim }}\,\frac{e^{x}-1}{x}.

    Hence \underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-2^{h}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{5^{h}-1}{h}-\frac{2^{h}-1}{h} \right).

    Now, consider \underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-1}{h}=\ln 5\cdot\underset{h\to 0}{\mathop{\lim }}\,\frac{e^{h\ln 5}-1}{h\ln 5}=\ln 5. In the same fashion we can find the value of the another limit, hence, both limits exist so we get that \underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-2^{h}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-1}{h}-\underset{h\to 0}{\mathop{\lim }}\,\frac{2^{h}-1}{h}=\ln \frac{5}{2}.
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