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Math Help - More Limits, please help.

  1. #1
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    More Limits, please help.

    There are around 70 question in my notebook, and the teacher ofcourse had to assign those without answers in the back. I know its a lot but any help would be appreciated.. I just can't seem to figure these limits.

    10.2

    Find the limit (if it doesn't exists, so state that, or use the symbols infinity and -infinity where appropriate.

    #12: lim h-->5- (from the left) √(5-h)

    my answer (pelase tell me if im correct, if not please explain)

    since when h approaches 5 from the left √(5-h) gets closer and closer to 0 .. therefore the limit is 0 ??

    16. lim x-->2+(from right) (x √ x^2-4) note** the x^2-4 i under the root

    as x gets closer to 2 from the right (x √ x^2-4) gets closer and closer to 0 ?

    22. lim x--> ∞ 2x-4 / 3-2x

    i read for limits at infinity you take the highest (dominant term) from the numerator and denominator

    so 2x / 2x = 1 therefore the limit is 1?

    34. lim x--> ∞ (4 - 3x^3) / (x^3-1)

    therefore by the same rule 3x^3 / x^3 = 3 therefore the limit is 3?

    44. lim x--> -2+ (from right) x / (√ 16-x^4) note** the 16-x^4 is under the root

    how would you do this? please explain.


    46. lim x --> ∞ ( x + 1/x)

    48. lim x--> 1/2 1 / 2x-1



    thank you very much for your help.. i know its a lot but i am much obliged.
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  2. #2
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    The first two look correct.

    For the next two, we must remember to account for differences in the signs of the terms in the numerator and denominator. For example,

    \frac{2x-4}{3-2x}=\frac{2-\frac{4}{x}}{-2+\frac{3}{x}}.

    Occasionally, limits do not exist as when the numerator approaches a nonzero value while the denominator approaches 0.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by alisheraz19 View Post
    There are around 70 question in my notebook, and the teacher ofcourse had to assign those without answers in the back. I know its a lot but any help would be appreciated.. I just can't seem to figure these limits.

    10.2

    Find the limit (if it doesn't exists, so state that, or use the symbols infinity and -infinity where appropriate.

    44. lim x--> -2+ (from right) x / (√ 16-x^4) note** the 16-x^4 is under the root

    how would you do this? please explain.


    46. lim x --> ∞ ( x + 1/x)

    48. lim x--> 1/2 1 / 2x-1



    thank you very much for your help.. i know its a lot but i am much obliged.
    44. As x\to-2, the denominator of that fraction gets very close to zero, so you end up with something like \frac{-2}{0}, so the limit is -\infty.

    46. Assuming that's \frac{x+1}{x}, use a similar technique that Scott showed you.

    48. What is the value of this if x=0.49? What if x=0.51? What does this tell you about the limit?
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  4. #4
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    44. lim x--> -2+ (from right) x / (√ 16-x^4) note** the 16-x^4 is under the root
    44. As , the denominator of that fraction gets very close to zero, so you end up with something like , so the limit is .
    Okay, I think I understand. So essentially can I plug in a number that is very close to -2 and then do it that way because it's hard for me (for some reason.. probably innate reasons) to see whether its - infinity or positive. So for example

    If I plug in a number close to -2 from the right such as -1.999 therefore

    -1.999 / (√ 16-(-1.999)^4) = a negative number therefore it is negative infinity.. ?

    can i do it like that?
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  5. #5
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    Quote Originally Posted by Scott H View Post
    The first two look correct.

    For the next two, we must remember to account for differences in the signs of the terms in the numerator and denominator. For example,

    \frac{2x-4}{3-2x}=\frac{2-\frac{4}{x}}{-2+\frac{3}{x}}.

    Occasionally, limits do not exist as when the numerator approaches a nonzero value while the denominator approaches 0.

    Hi Scott, thank you for your answers but I have to say I'm not quite following you came to that answer.... i mean.. no.. wait I understand how you came to it because but I don't understand why.. It's strange because in my book this is what it says about Limits at infinity:

    here's the example.. Lim x--> infinity

    (4x^2 + 5) / (2x^2+1)

    ||from the book|| For the preceding function there is an easier way to find the lim x--> infinity

    For large values of x, in the numerator the term involving the greatest power of x, namely 4x^2, dominates the sum 4x^2+5, and the dominant term in the denominator 2x^2 +1 is 2x^2. Thus as x--> infinity, f(x) can be approximated by (4x^2)/ (2x^2)

    thus: Lim x-->infinity 4x^2 / 2x^2 = lim x --> infinity 2 = 2

    why can't this be applied to my question:

    22. lim x--> ∞ 2x-4 / 3-2x

    therefore 2x/2x = 1?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by alisheraz19 View Post
    Okay, I think I understand. So essentially can I plug in a number that is very close to -2 and then do it that way because it's hard for me (for some reason.. probably innate reasons) to see whether its - infinity or positive. So for example

    If I plug in a number close to -2 from the right such as -1.999 therefore

    -1.999 / (√ 16-(-1.999)^4) = a negative number therefore it is negative infinity.. ?

    can i do it like that?
    Numerical approximation works most of the time, but once in a blue moon you'll get one that's deceptive...I can't think of any right now.

    Quote Originally Posted by alisheraz19 View Post
    Hi Scott, thank you for your answers but I have to say I'm not quite following you came to that answer.... i mean.. no.. wait I understand how you came to it because but I don't understand why.. It's strange because in my book this is what it says about Limits at infinity:

    here's the example.. Lim x--> infinity

    (4x^2 + 5) / (2x^2+1)

    ||from the book|| For the preceding function there is an easier way to find the lim x--> infinity

    For large values of x, in the numerator the term involving the greatest power of x, namely 4x^2, dominates the sum 4x^2+5, and the dominant term in the denominator 2x^2 +1 is 2x^2. Thus as x--> infinity, f(x) can be approximated by (4x^2)/ (2x^2)

    thus: Lim x-->infinity 4x^2 / 2x^2 = lim x --> infinity 2 = 2

    why can't this be applied to my question:

    22. lim x--> ∞ 2x-4 / 3-2x

    therefore 2x/2x = 1?
    It can, but the coefficient of x in the denominator is -2, so you end up with \lim_{x\to0}\frac{2x}{-2x}=-1
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    Numerical approximation works most of the time, but once in a blue moon you'll get one that's deceptive...I can't think of any right now.
    I would say that if we let \sigma_n=\sum_{m=1}^n\frac{1}{m} then taking bigger and bigger values for n in \sigma_n surely does not lead one to believe that \lim\sigma_n\to\infty
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  8. #8
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    Quote Originally Posted by alisheraz19 View Post
    ||from the book|| For the preceding function there is an easier way to find the lim x--> infinity

    For large values of x, in the numerator the term involving the greatest power of x, namely 4x^2, dominates the sum 4x^2+5, and the dominant term in the denominator 2x^2 +1 is 2x^2. Thus as x--> infinity, f(x) can be approximated by (4x^2)/ (2x^2)

    thus: Lim x-->infinity 4x^2 / 2x^2 = lim x --> infinity 2 = 2
    The book's wording is, in my opinion, rather vague, but the idea is that dividing the numerator and denominator of a fraction by the same number leaves the value of the fraction unchanged. Therefore,

    \frac{2x-4}{3-2x}=\frac{2x-4}{3-2x}\cdot 1=\frac{2x-4}{3-2x}\cdot\frac{1/x}{1/x}=\frac{\frac{2x}{x}-\frac{4}{x}}{\frac{3}{x}-\frac{2x}{x}}=\frac{2-\frac{4}{x}}{-2+\frac{3}{x}}.

    As x approaches \infty, the fractions with x in the denominator approach 0, leaving the limit expression to approach \frac{2}{-2}=-1.

    why can't this be applied to my question:

    22. lim x--> ∞ 2x-4 / 3-2x

    therefore 2x/2x = 1?
    You are almost correct. We merely need to account for the sign of -2x in the denominator. The correct answer is

    \frac{2x}{-2x}=-1.
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  9. #9
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    Thank you very much everyone, that was unbelievably quick. I appreciate everyone's help and effort into answering my questions. Rest assured there will be more to come later

    thank you again!
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