limits question (1 Viewer)

[edd91]

can you assess the limit x->0 of cosx/x
or of 1-cosx/x ?

 

[Xayma]

as x-->0 cos x-->1

ie limit (x-->0) cosx/x
=limit (x-->0) 1/x
=∞

 

[CM_Tutor]

lim_x-->0⁺ cos x / x = +∞
lim_x-->0^- cos x / x = -∞
lim_x-->0 (1 - cos x) / x = 0

 

[kpq_sniper017]

Originally posted by CM_Tutor
lim_x-->0⁺ cos x / x = +∞
lim_x-->0^- cos x / x = -∞
lim_x-->0 (1 - cos x) / x = 0

do we need to have a knowledge of limits other than sinx/x for the hsc exam?

 

[Xayma]

You need to understand small angles which are limits in themselves.

Do those first then it becomes an ordinary limit.

 

[CM_Tutor]

Originally posted by pcx_demolition017
do we need to have a knowledge of limits other than sinx/x for the hsc exam?

You need to be able to figure out simple limits. The first of these has cos x, which is 1 at x = 0, and so we have a 1 / x - hence, +/- ∞, depending on the sign of x.

For the last one, rewrite 1 - cos x as a 2sin²x / 2. It follows that (1 - cos x) / x = sin(x / 2) * sin(x / 2) / (x / 2). The first term, sin(x / 2), goes to zero, and the second, sin(x / 2) / (x / 2). goes to 1. So, the limit is 1 * 0, using the 'typical' sin x / x limit.

 

[CM_Tutor]

I don't know, but I wouldn't run the risk if there is another way to get at the limit.

 

[CrashOveride]

My 2unit teacher said as long as you state it, it's okay (even tho its not in the syllabus - he said he has seen it before)

 

[kpq_sniper017]

Will we lose marks if we use L'Hopital's?

what exactly is l'hopital's method?
i've heard it in these posts, but i don't know what it is.
is it part of university maths??

 

[gman03]

what exactly is l'hopital's method?
i've heard it in these posts, but i don't know what it is.
is it part of university maths??

Yes it is a way of finding a limit at a point with an intermedia form of 0/0 or infinity/infinity

 

[Estel]

Explain?

 

[gman03]

Explain?

Quote from my notes:

Suppose functions f and g are differentiable, except possibly at a.

If both f(x) -> 0 and g(x) -> 0 as x -> a and
f'(x)/g'(x) -> L as x -> a,
then f(x)/g(x)->L as x -> a

Get it?

 

[CM_Tutor]

Applying L'Hopital's Rule to the question of interest here, we have: lim_x-->0(1 - cos x) / x

Direct substitution gives 0 / 0, so L'Hoptial's Rule is valid in this case, as the limit is of the correct form, and both the functions 1 - cos x and x are continuous and differentiable.

So, lim_x-->0(1 - cos x) / x
= lim_x-->0[d/dx(1 - cos x) / d/dx(x)]
= lim_x-->0[(0 - -sin x) / 1]
= lim_x-->0sin x
= sin 0
= 0

Note, however, that L'Hopital's Rule could not be used on the first problem presented, lim_x-->0cos x / x, as cos x / x is not 0 / 0 or +/- ∞ / ∞ at x = 0. Applying L'Hopital's Rule would give lim_x-->0cos x / x as 0, which is incorrect.

 

[Estel]

Yup.

Thankyou.

 

[kpq_sniper017]

Applying L'Hopital's Rule to the question of interest here, we have: lim_x-->0(1 - cos x) / x

Direct substitution gives 0 / 0, so L'Hoptial's Rule is valid in this case, as the limit is of the correct form, and both the functions 1 - cos x and x are continuous and differentiable.

So, lim_x-->0(1 - cos x) / x
= lim_x-->0[d/dx(1 - cos x) / d/dx(x)]
= lim_x-->0[(0 - -sin x) / 1]
= lim_x-->0sin x
= sin 0
= 0

Note, however, that L'Hopital's Rule could not be used on the first problem presented, lim_x-->0cos x / x, as cos x / x is not 0 / 0 or +/- ∞ / ∞ at x = 0. Applying L'Hopital's Rule would give lim_x-->0cos x / x as 0, which is incorrect.

so u could use l'hopital's for lim_x-->0 tanx/x?

 

[CM_Tutor]

so u could use l'hopital's for lim_x-->0 tanx/x?

Yes, as both tan x and x are continuous and differentiable in the region around x = 0, and at x = 0 we have 0 / 0.

 

[kpq_sniper017]

Yes, as both tan x and x are continuous and differentiable in the region around x = 0, and at x = 0 we have 0 / 0.

so basically, as long as each of the numerator and denominator are differentiable and give 0/0 when substituting x=0, u can use l'hopital's?
does it also work for lim_x-->a tanx/x, for any real number a?

 

[CM_Tutor]

so basically, as long as each of the numerator and denominator are differentiable and give 0/0 when substituting x=0, u can use l'hopital's?
does it also work for lim_x-->a tanx/x, for any real number a?

It won't work for lim_x-->a tan x / x unless a = 0, as this is the only place you can get a 0 / 0 result. However, L'Hopital's Rule will work for any case where a 0 / 0 results. So, you could use it for

lim_x-->5 [sqrt(30 - x) - 5] / (5 - x)

It's up to you to decide whether this is faster than the standard Extn 1 solution to this problem.

 

[kpq_sniper017]

It won't work for lim_x-->a tan x / x unless a = 0, as this is the only place you can get a 0 / 0 result. However, L'Hopital's Rule will work for any case where a 0 / 0 results. So, you could use it for

lim_x-->5 [sqrt(30 - x) - 5] / (5 - x)

It's up to you to decide whether this is faster than the standard Extn 1 solution to this problem.

but referring to the original post....we might possibly get marked down if we do?

 

[CM_Tutor]

but referring to the original post....we might possibly get marked down if we do?

You might - it depends on the marker. However, if you don't know any other way to do the problem, then getting the answer using L'Hopital's Rule is much better than leaving the page blank.