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June 1st, 2017, 01:52 PM   #1
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limit involving trig

Hi Guys,

I am trying to find the best way to solve this limit. I know that it is undefined.

$$\lim_{x\to\infty}\frac{\tan 8x}{x+\sin 3x}$$

I tried L'Hôpital, but am getting stuck.

Thanks.

Last edited by skipjack; June 1st, 2017 at 09:54 PM.
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June 1st, 2017, 02:23 PM   #2
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With L'Hôpital

$$\lim_{x\to\infty}\frac{8\sec^{2}8x}{1+3\cos3x}$$

Is it ok to stop here and state that the numerator oscillates between infinity and negative infinity and that the denominator is trapped between 2 and 0 and conclude that the limit is undefined?

I need to put that in math terms.

Last edited by skipjack; June 1st, 2017 at 09:55 PM.
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June 1st, 2017, 02:26 PM   #3
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Proof by Wolfram Alpha. Oscillates.

https://www.wolframalpha.com/input/?...x+%2B+sin(3x))
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June 1st, 2017, 02:35 PM   #4
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Quote:
Originally Posted by Maschke View Post
Proof by Wolfram Alpha. Oscillates.

https://www.wolframalpha.com/input/?...x+%2B+sin(3x))
I am not sure how to find the limit from there. It looks like it just gave me other forms of the expression.

Last edited by skipjack; June 1st, 2017 at 09:55 PM.
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June 1st, 2017, 02:42 PM   #5
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Quote:
Originally Posted by dthiaw View Post
am not sure how to find the limit from there. it looks like it just gave me other forms of the expression.
The graph shows that it oscillates as suspected. Of course that's not a proof.
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June 1st, 2017, 02:44 PM   #6
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Quote:
Originally Posted by Maschke View Post
The graph shows that it oscillates as suspected. Of course that's not a proof.
Yeah, but I wanted to see the math. I am almost certain that L'Hôpital is involved.

Last edited by skipjack; June 1st, 2017 at 09:56 PM.
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June 1st, 2017, 03:49 PM   #7
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Quote:
Originally Posted by dthiaw View Post
Yeah, but I wanted to see the math. I am almost certain that L'Hôpital is involved.
$\dfrac{\tan(8x)}{x+\sin{x}}$ doesn't satisfy the hypotheses of L'Hôpital's Rule, which require that the limits in numerator and denominator either both
exist and be zero or both be infinite.

Have a look at the link which discusses a similar limit ...

https://math.stackexchange.com/quest...does-not-exist
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Last edited by skipjack; June 1st, 2017 at 09:57 PM.
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June 1st, 2017, 03:55 PM   #8
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Quote:
Originally Posted by dthiaw View Post
Hi Guys,

I am trying to find the best way to solve this limit. I know that it is undefined.

$$\lim_{x\to\infty}\frac{\tan 8x}{x+\sin 3x}$$

I tried L'Hôpital, but am getting stuck.

Thanks.
When $8x = 2n\pi + \frac\pi2$, the numerator is undefined, switching from arbitrarily large and positive to arbitrarily large and negative. At this moment, the denominator is positive and finite for large $x$ and so the value of the function switches from arbitrarily large and positive to arbitrarily large and negative. Thus there can be no limit.
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Last edited by skipjack; June 1st, 2017 at 09:58 PM.
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June 1st, 2017, 04:02 PM   #9
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Thanks a bunch, guys. It's finally clear.

Last edited by skipjack; June 1st, 2017 at 09:59 PM.
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June 1st, 2017, 06:40 PM   #10
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Thinking about it, the switching from positive to negative isn't necessary. Simply the fact that for any given value of $x$ close to $\frac18(2n+\frac12)\pi$ the fraction becomes arbitrarily large is enough to deny convergence.

Last edited by skipjack; June 1st, 2017 at 09:59 PM.
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