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June 11th, 2014, 06:37 PM   #1
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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.
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June 11th, 2014, 08:41 PM   #2
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Are you sure you're differentiating properly? I'm getting the same answer both ways.

PS: $\frac{d}{dx}x = 1$
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June 11th, 2014, 09:48 PM   #3
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lim (x-cosx)/x .sorry made a mistake.
x->0
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June 11th, 2014, 09:49 PM   #4
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if you differentiate (1-sinx)/1.
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June 12th, 2014, 02:21 PM   #5
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At zero.
Quote:
Originally Posted by harryjobs View Post
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.
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June 12th, 2014, 03:35 PM   #6
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Quote:
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lim (x-cosx)/x .sorry made a mistake.
x->0
L'Hopital's rule does not apply. Numerator -> -1 and denominator -> 0.
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June 13th, 2014, 08:35 PM   #7
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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
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June 13th, 2014, 09:25 PM   #8
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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.
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