My Math Forum L-hospitals rule showing limit does not exist.?

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 June 11th, 2014, 06:37 PM #1 Newbie   Joined: Jun 2014 From: chennai Posts: 17 Thanks: 0 L-hospitals rule showing limit does not exist.? lim x-cosx/x . why cant we use L hospitals rule.if we I use it I am getting limit does x->infinity not exist but if i divide by x then apply limit i am getting one.ones the answer but at the same time limit does not exist what does this mean.
 June 11th, 2014, 08:41 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Are you sure you're differentiating properly? I'm getting the same answer both ways. PS: $\frac{d}{dx}x = 1$
 June 11th, 2014, 09:48 PM #3 Newbie   Joined: Jun 2014 From: chennai Posts: 17 Thanks: 0 lim (x-cosx)/x .sorry made a mistake. x->0
 June 11th, 2014, 09:49 PM #4 Newbie   Joined: Jun 2014 From: chennai Posts: 17 Thanks: 0 if you differentiate (1-sinx)/1.
June 12th, 2014, 02:21 PM   #5
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At zero.
Quote:
 Originally Posted by harryjobs lim (x-cosx)/x .sorry made a mistake. x->0
That's not an indeterminate form, so l'Hopitall doesn't apply. It just blows up at zero.

For the limit at infinity, the same applies to the cosine term. It's not indeterminate.

Last edited by v8archie; June 12th, 2014 at 02:27 PM.

June 12th, 2014, 03:35 PM   #6
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Quote:
 Originally Posted by harryjobs lim (x-cosx)/x .sorry made a mistake. x->0
L'Hopital's rule does not apply. Numerator -> -1 and denominator -> 0.

June 13th, 2014, 08:35 PM   #7
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Quote:
 Originally Posted by v8archie At zero. That's not an indeterminate form, so l'Hopitall doesn't apply. It just blows up at zero. For the limit at infinity, the same applies to the cosine term. It's not indeterminate.
why doesnt it apply thats what i want to know

 June 13th, 2014, 09:25 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Firstly, there's nothing it can do to help. If the form is not indeterminate, that means we know the value. In this case, we know that it grows without bound as we near the limit. It can't do anything else. Indeterminate forms are particular combinations of $0$ and $\pm\infty$ that can, under different conditions produce different values. Thus $$\lim_{x \to \infty} \frac{x -1}{x} = 1 \qquad \lim_{x \to \infty} \frac{x -1}{x^2} = 0 \qquad \lim_{x \to \infty} \frac{x^2 -1}{x} = \infty$$ In each case, the quotient is heading towards $\frac{\infty}{\infty}$, but we see three different results depending on the relative speed that the numerator and denominator grow. Thanks from harryjobs

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