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Limits and Derivatives

  1. Oct 5, 2005 #1

    (1)[tex] \lim_{x\rightarrow -\infty} \frac{x-2}{x^{2} + 2x + 1} [/tex]. I factored it as [tex] \frac{x-2}{(x+1)^{2}} [/tex]. Then what?
    (2) [tex] \lim_{x\rightarrow -\infty} \frac{\sqrt{5x^{2}-2}}{x+3} [/tex]. For this one would I just multiply both sides by the numerator? I am not sure what to do with this one.
    (3) [tex] \lim_{x\rightarrow -\infty} \frac{\sqrt{3x^{4}+x}}{x^{2}-8} [/tex]. Would I do the same thing and multiply both sides by the numerator?
    (4) [tex] \lim_{x\rightarrow 3} \frac{x}{x-3} [/tex]. Is there any way I can separate this?
    (5) [tex] \lim_{x\rightarrow 4-} \frac{3-x}{x^{2}-2x-8} [/tex]. Would I just factor both the numerator and denominator?
    (6) [tex] \lim_{x\rightarrow\infty} \frac{7-6x^{5}}{x+3} [/tex]. For this one would I also factor? Not sure how to do it.
    (7)[tex] \lim_{x\rightarrow 0-} \frac{x}{|x|} [/tex]. This would just be -1?
    (8) [tex] \lim_{x\rightarrow 0} \frac{\sin 2\theta}{\theta^{2}} [/tex]. This wouldnt exist? [tex] \frac{\sin 2\theta}{\theta^{2}} = 2\cos\theta(\frac{\sin\theta}{\theta})(\frac{1}{\theta}) [/tex]. How would I show this algebraically?

    Last edited: Oct 5, 2005
  2. jcsd
  3. Oct 5, 2005 #2
    Alrighty, ill help you with as much as i have time for right now.
    For the first 3, what you need to do is do divide top and bottom by your highest power of x in the denominator. ill show you the first one and you can do the other 2.

    if i divide by x^2 on top and bottom, i get
    (1/x) - (2/x^2) divided by (1 + 2/x + 1/x^2)

    now, your x is tending to negative infinity, but here the sign is irrelevant. think about it. if i divided a constant by a number thats getting bigger and bigger, itll become closer to 0 right? so replace any instance of (1/x) and those things, with 0.
    so your numerator will tend towards 0. and therefore your entire limit will be 0.

    in 2 and 3 its the same concept, just you need to square the power of x and you divided by in the numberator to be able to insert it into the root. the minus infinity will thus become important because you might need a sign change. (because -infinity= -root of infinity squared).
    try and work it out and tell me what you find.

    for the 4th one, you cant factor it. so you must so a 2 case workout of the possibilities. one where you approach the limit from the left, and another where you approach it from the right and you compare the two limits and see if they are equal. if they are not, you can say that the limit does not exist. (when i say left, i mean negative) so replace x by something very very close (like, +0.000001 difference) and another time with ( -0.000001) difference, and see what you notice.

    5) the full limit doesnt exist, even if you do factor it, but it says to approach x as 3-. that means from the negative side. and see what kind of infinity you get.

    thats all i have time for, please write back if you have questions so far. i didnt really want to give you the answers so much as give you something to think about.

    have fun!
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