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Homework Help: Limit's Help work shown please help!

  1. Feb 17, 2009 #1

    a.) lim x-->infinity {(3x^3)+ cos(x)} / {(sin(x)- x^3)}

    b.) lim x-->pi(+) {(tan^-1 (1/(x-pi))} / {pi-x0}

    c.) lim x--> 0(+) {sqrt(x+ sin(x))}*(ln x)

    d.) lim x--> 1(-) {(cos^-1 (x))} / (1-x)

    My work process:

    a.) I divided the numerator and the denominator by the highest power which is 3... therefore
    the expression becomes {3+ cos(x)/3}/ {sin (x)/3 - 1)}
    since, 1/infinity becomes zero
    the limit should be -3

    b.)for the numerator of the expression as x approaches pi from the positive side the inverse of tangent is 1/0 which is undefined... so the expression of the numerator is undefined..... also... as x approaches pi from the positive side the denominator of the expression approaches 0.. so the overall expression becomes 1/0... therefore the limit does not exist.. and since it is not bound... i'm sort of iffy on my answer for this one in terms of my reasoning and evaluation please help me out

    c.) as x approaches zero from the positive side, the square root of the expression x+ sin(x) becomes zero, however the natural logarithm of zero is undefined... therefore the product of 0 and an undefined quantity doesn't produce a limit... therefore the limit is undefined... i'm sort of iffy on my answer for this question as well

    d.) as approaches 1 from the negative side the numerator of the expression of inverse cosine of x becomes zero, the denominator also becomes zero since the expression of the denominator is 1-x.... therefore the limit is zero for this expression.. i'm sort of iffy for this expression as well

    Please help me with these problems... i have provided some of my work process
  2. jcsd
  3. Feb 17, 2009 #2


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    Science Advisor

    There is no 1/infinity in what you wrote! In any case, you don't divide by "the highest power", you divide by the term with the highest power. You should have
    [tex]\frac{3+ \frac{cos(x)}{x^3}}{\frac{sin(x)}{x^3}- 1}[/tex]
    Now, that has an x in the denominator which goes to 0 and so the limit is -3. Is that what you meant?

    [itex]tan(\theta)[/itex] goes to infinity as [itex]\theta[/itex] goes to [itex]\pi/2[/itex] so [itex]tan^{-1}(x)[/itex] goes to [itex]\pi/2[/itex] as x goes to infinity. The limit of the expression in the numerator is NOT undefined. But the fact that the limit in the numerator is a non-zero number and the denominator goes to 0 does mean the limit does not exist.

    You are right that the [itex]\sqrt{x+ sin(x)}[/itex] goes to 0 and that ln(x) goes to negative infinity as x goes to 0. However, that does not mean the limit does not exist. For example (x)(-1/x)= -1 even though (x) goes to 0, and (-1/x) goes to negative infinity. "[itex]0\cdot\infnty[/itex]" is "undetermined". I think you are going to L'Hopital's rule for this.

    No, no, no! 0/0 is not 0, it is another "undetermined" limit. Use L'Hopital's rule.

    Please help me with these problems... i have provided some of my work process[/QUOTE]
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