Vividness
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#1
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Can anyone help with this question? Im not sure what method to use...

Find the limit of ((x^2)+sinx)/(x^2)

as x approaches infinity. Thanks
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ccfcadam36
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If you split the fraction up into (x^2)/(x^2)+(sinx)/(x^2), then thi should make the question easier
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Hasufel
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use \displaystyle \frac{|\sin(x)|}{x^2} with the Pinching theorem - adding 1 to the inequality - the answer is pretty obvious

(EDIT): previous poster`s post is easier, though!
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Vividness
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(Original post by ccfcadam36)
If you split the fraction up into (x^2)/(x^2)+(sinx)/(x^2), then thi should make the question easier
Okay, so im trying to find limit of 1+(sinx)/(x^2), I know sinx will be between -1 and 1, and x^2 will go to infinity, so can i just say that (sinx)/(x^2) tends to 0, so the limit is 1? And is there a special rule for finding (sinx)/(x^2)?
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Hasufel
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with thr Pinching theorem (as above) you can show that - because you take the modulus of Sin(x) (since it doesn`t deviate outside +/-1):

\displaystyle \frac{-1}{x^2} \leq \frac{ \sin(x)}{x^{2}} \leq \frac{1}{x^2}

so you get, adding the "1":

\displaystyle 1+\frac{-1}{x^2} \leq 1+\frac{ \sin(x)}{x^{2}} \leq 1+ \frac{1}{x^2}
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ccfcadam36
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(Original post by Vividness)
Okay, so im trying to find limit of 1+(sinx)/(x^2), I know sinx will be between -1 and 1, and x^2 will go to infinity, so can i just say that (sinx)/(x^2) tends to 0, so the limit is 1? And is there a special rule for finding (sinx)/(x^2)?
What you've said is correct, and there is no special rule as far as I'm aware, web dealing with fractions it's just best to simplify them as much as possible and then consider the top and bottom separately to find the overall limit, hope this helps!
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Vividness
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Thanks for your help on that one, just one more I'm stuck on:

limit of (sqrt(9x+1))/(x+1) as x approaches infinity, any ideas?
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ccfcadam36
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(Original post by Vividness)
Thanks for your help on that one, just one more I'm stuck on:

limit of (sqrt(9x+1))/(x+1) as x approaches infinity, any ideas?
Firstly,Consider the limit of (9x+1)/(x+1) by dividing everything by x to give (9+1/x)/(1+1/x), then if you know the limit of 1/x as x tend to infinity then you can work out the limit of the whole thing
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Vividness
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(Original post by ccfcadam36)
Firstly,Consider the limit of (9x+1)/(x+1) by dividing everything by x to give (9+1/x)/(1+1/x), then if you know the limit of 1/x as x tend to infinity then you can work out the limit of the whole thing
So without the square root the limit would be 9, but how can i adapt this solution to find the limit when the numerator is square rooted?
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ccfcadam36
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(Original post by Vividness)
So without the square root the limit would be 9, but how can i adapt this solution to find the limit when the numerator is square rooted?
Sorry I misread, I thought the whole thing was square rooted, using the same method of dividing everything by x will still work it just means the numerator will be sqrt(9+1/x) and the denominator will be simply 1+1/x
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Vividness
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#11
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(Original post by Hasufel)
with thr Pinching theorem (as above) you can show that - because you take the modulus of Sin(x) (since it doesn`t deviate outside +/-1):

\displaystyle \frac{-1}{x^2} \leq \frac{ \sin(x)}{x^{2}} \leq \frac{1}{x^2}

so you get, adding the "1":

\displaystyle 1+\frac{-1}{x^2} \leq 1+\frac{ \sin(x)}{x^{2}} \leq 1+ \frac{1}{x^2}
Thanks for that, any ideas on this one?
limit of (sqrt(9x+1))/(x+1) as x approaches infinity
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davros
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(Original post by Vividness)
So without the square root the limit would be 9, but how can i adapt this solution to find the limit when the numerator is square rooted?
Intuitively you should be able to see that this limit is 0 too because the denominator gets much bigger than the numerator as x goes to infinity.

You can still divide top and bottom by x but note that you have to take factor 1/x^2 inside the square root.

Alternatively, you can make some "brute force" approximations - x+1 > x so 1/(x+1) < 1/x gives you a simpler thing to work with, and then use the squeeze theorem to put the limit between 2 bounds.
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Hasufel
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(Original post by Vividness)
Thanks for that, any ideas on this one?
limit of (sqrt(9x+1))/(x+1) as x approaches infinity
One way would be to first multiply top and bottom by (9x+1)^{1/2}

This gets rid of the root on the numerator - moving it to the denominator where it`s easier to eliminate.

we then end up with:

\displaystyle (\frac{9x+1}{x+1})(\frac{1}{9x+1})^{1/2}

splitting up the first part, we get:

\displaystyle (9- \frac{8}{x+1})(\frac{1}{9x+1})^{1/2}

Now take limits (Lim_{x-&gt;a}\sqrt{f(x)}=\sqrt{Lim_{x-&gt;a}f(x)} )
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DFranklin
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(Original post by Vividness)
Thanks for your help on that one, just one more I'm stuck on:

limit of (sqrt(9x+1))/(x+1) as x approaches infinity, any ideas?
Note that 0 &lt; \dfrac{\sqrt{9x+1}}{x+1} &lt; \dfrac{\sqrt{9(x+1)}}{x+1} = \dfrac{3}{\sqrt{x+1}}. The final limit on the RHS is hopefully straightforward.
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