Trig Limits Question (1 Viewer)

[Studentleader]

Lim x -> 0 (cosx(1+2sinx))/x - 1/x

Few questions like this which I just can't get, help would be appreciated.

 

[solomarc20]

Is the answer 0?

Lim x-> 0 (cos x(1+2sinx))/ 1- (1/x)

= Lim x-> 0 (cos x+ 2sinxcosx)/ (x-1)/x
= Lim x-> 0 x(cosx+ sin2X)/x-1
= 0

I'd double check this, as I'm not entirely sure...

 

[namburger]

my calc says 2

Lim x -> 0 (cosx(1+2sinx))/x - 1/x
Here are identities:
Lim x -> 0 (Sin x)/x = 1
Lim (1 - cos x)/x = 0

Lim x -> 0 (cosx(1+2sinx))/x - 1/x)
= Lim x -> 0 (1/x) (cos x + Sin 2x - 1)
You can split lims
= Lim x -> 0 (Sin 2x)/x + Lim x -> 0 (cos x - 1)/x
= Lim x -> 0 2(Sin 2x)/2x - Lim x -> 0 (1 - cos x)/x
= 2

 

[Slidey]

Lim x -> 0 (cosx(1+2sinx))/x - 1/x

Few questions like this which I just can't get, help would be appreciated.

Lim x -> 0 (cosx(1+2sinx))/x - 1/x =
Lim x -> 0 (cosx(1+2sinx) - 1)/x =

Since both the top and bottom go to 0 and are continuous, use L'Hopital's rule (take derivative of top and bottom, then try the limit again):
Lim x -> 0 (cosx(1+2sinx) - 1)/x =
Lim x -> 0 (-sinx+2cos2x))/1 = 2

 

[pLuvia]

Lim x -> 0 (cosx(1+2sinx))/x - 1/x =
Lim x -> 0 (cosx(1+2sinx) - 1)/x =

Since both the top and bottom go to 0 and are continuous, use L'Hopital's rule (take derivative of top and bottom, then try the limit again):
Lim x -> 0 (cosx(1+2sinx) - 1)/x =
Lim x -> 0 (-sinx+2cos2x))/1 = 2

lol that's what I was going to do but unfortunately L'Hopital's rule isn't taught in the MX2 course

 

[namburger]

Since both the top and bottom go to 0 and are continuous, use L'Hopital's rule (take derivative of top and bottom, then try the limit again):

What do you mean by top and bottom go to 0. is it if you sub x = 0 in or what.
What do you mean by continuous?
Can i use this method in my exam?

 

[pLuvia]

What do you mean by top and bottom go to 0. is it if you sub x = 0 in or what.
What do you mean by continuous?
Can i use this method in my exam?

If you want to use L'Hopital's rule, if the numerator and the denominator tend to zero as x approaches 0 then you differentiate both the numerator and the denominator then you do the limit again and you will get the answer

I'm not sure if you are allowed to use this rule in HSC exams because you learn this method in 1st year undergraduate maths

 

[Slidey]

There's always another way to solve limits, so try that first. If it fails, use L'Hopital's and you'll probably solve it in 2 seconds. You'll then be awarded full marks and aplauded for your ingenuity.

They are only allowed to deduct marks when they say "hence, solve" and you don't use the hence, or if they say "using method X, solve" and you don't use method X.

If they say "hence or otherwise", you may freely use whatever method you wish.

Conditions: L'Hopital's rule only works if:
a) both the numerator and denominator tend towards the same value when you consider their limits individually.
b) that "same value" must be an indeterminate form: 0, infinity, negative infinity, 0^0, 0*infinity, etc.
c) the derivative of the top and bottom exist.

Sometimes the derivatives just oscillate around a bit (e.g. sometimes with trig and exponentials) and thus the rule isn't useful.

You should really know what continuity is by now:

http://en.wikipedia.org/wiki/Continuous_function

 

[solomarc20]

Whoops... I thought the 1/x was a part of the denominator... God I can b really stupid at times!!!!

 

[Slidey]

Nah it's a fair enough mistake. A lot of people come on here and post things without brackets. I just figured that because he'd put so much effort into brackets for the numerator, he wouldn't just suddenly forget on the denominator.

 

[kevinant]

l'Hopital rule isn't taught in 4U though but it's a fairly easy concept (It's in my assignment)
Lim f(x)/g(x) =
x->0

Lim f'(x)/g'(x) when
x->0

Lim x->0 f(x) and lim x->0 g(x) is either 0 or infinity.
This works because the slope indicates the rate of the limit approaching to 0 or inifinity. Hence the limit is workable.


However, that's not the smartest thing to do..the smartest thing to do is USE YOUR CALCULATOR... that's what I did when i see a weird limit question.

just plug the formula into your calculator (replace all x with "ans") and then pick a number for ans which is close to 0 in this case.

Wow there are two other people from my school posted in this thread.

 

[Slidey]

Using your calculator won't help most of the time since the markers look for a method used.