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    Calculate lim x-> infinity for xn = (sqrtn)(n+e-n)-1(sin(e)n)

    Please can anyone help/give me a hint? I've been trying to find the limit for ages with loads of different methods.. I thought squeeze theorem but it doesn't look to be working. Then I tried to get n/e^n somewhere because that's a standard result.. no luck. Gah

    Thanks for any suggestions xx
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    (Original post by JohnnySPal)
    So that's \lim_{n\rightarrow \infty} \frac{\sqrt{n}}{n+e^{-n}}(\sin{e})^n?
    No sorry =/ It's this:

    \lim_{n\rightarrow \infty} \frac{\sqrt{n}}{n+e^{-n}}\sin({e}^n)

    I'm not great with latex!
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    (Original post by JohnnySPal)
    So that's \lim_{n\rightarrow \infty} \frac{\sqrt{n}}{n+e^{-n}}(\sin{e^n})

    It looks like a classic case of working with mods and noticing that 0\leq |\sin{x}|\leq 1 \forall x
    Hmm ok.. I did try that a little but then I just got stuck with the e on the denom. I'm so tired =( and limits are so new to me.. please could you give me a hint?
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    Well, you could always write:

    \frac{\sqrt{n}}{n + e^{-n}} = \frac{1}{\sqrt{n} + \frac{e^{-n}}{\sqrt{n}} }

    after using the sandwich rule. Then you'll have 0 <= Your crap <= something that goes to 0. Does this help?

    I think this works anyway :erk:
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    (Original post by JohnnySPal)
    Well, you could always write:

    \frac{\sqrt{n}}{n + e^{-n}} = \frac{1}{\sqrt{n} + \frac{e^{-n}}{\sqrt{n}} }

    after using the sandwich rule. Then you'll have 0 <= Your crap <= something that goes to 0. Does this help?

    I think this works anyway :erk:
    Does the bit with the e tend to 0 now? Because it's like:

    \frac{1}{\sqrt{n} + \frac{1/\sqrt{n}}{e^{n}} }


    Edit: Ohh sandwich rule.. so I say that the e^-n > 0 and then if you just have 1 over sqrt n that'll be bigger than my crap?
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    (Original post by TheRandomer)
    Does the bit with the e tend to 0 now? Because it's like:

    \frac{1}{\sqrt{n} + \frac{1/\sqrt{n}}{e^{n}} }
    Either way you look at it, the fraction on the denominator eventaully becomes like 0/infinity, so it definitely goes to 0.

    I whacked your x_n into my graphical calculator and it looks like it goes to zero as n goes off to infinity, albeit rather chaotically (informal use of the term, maths nerds :p:).


    EDIT: Well, using the fact that 0 <= |your thing| <= |this fraction we've been discussing| = this fraction we've been discussing again (it's a positive function so just get rid of the mod signs), along with the fact that this fraction we've been discussing goes off to 0 proves that your function goes off to 0 by the sandwich rule. If that's what you meant.
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    (Original post by JohnnySPal)
    Either way you look at it, the fraction on the denominator eventaully becomes like 0/infinity, so it definitely goes to 0.

    I whacked your x_n into my graphical calculator and it looks like it goes to zero as n goes off to infinity, albeit rather chaotically (informal use of the term, maths nerds :p:).
    Thankyouuu =) *reprep*

    God I hate analysis right now. Even if it's really obvious what will happen it's all this proving and disproving - it really annoys me!
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    (Original post by TheRandomer)
    Thankyouuu =) *reprep*

    God I hate analysis right now. Even if it's really obvious what will happen it's all this proving and disproving - it really annoys me!
    I usually find it's best if I think about whatever sequence a_n you're considering as a function of x and to whack it into a graphical calculator (or online graph drawer). At least that way you know for certain what limit you're aiming for before you start applying about a million techniques to it.
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    Except you don't get graphical calculators in exams. The easiest way to do this question is the sandwich rule:

    \frac{\sqrt{n} sin(e^n)}{n + \frac{1}{e^n}} \leq \frac{\sqrt{n}}{n + \frac{1}{e^n}} &lt; \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}} which tends to zero as n goes to infinity. You can do a similar thing with the left hand-side(or just use zero) to get the "sandwich"
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    Yeahh Barny has a slightly better sandwich, but you still need to throw mod signs into that working out.

    If you're doing assignment sheet questions in yur own time graphical calculators can be realy helpful. For this level of work you can build up a great intuition after trying a load of example, so you can gradually use graphical aids less and less. It certainly helped me during my first year, anyway
 
 
 
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