The Student Room Group

calculus limit q

I got this far... trying to calculate the limit of the first thing I wrote

Posted from TSR Mobile
Reply 1
actually I think I got it.

Posted from TSR Mobile
Original post by cooldudeman
actually I think I got it.

Posted from TSR Mobile


Note that

tanxsinxx3=sin3xx3×1cosx(cosx+1)\dfrac{\tan x - \sin x }{x^3} = \dfrac{\sin^3 x }{x^3} \times \dfrac{1}{\cos x (\cos x+1)}

Can you see where you might have gone wrong?
Reply 3
Original post by Indeterminate
Note that

tanxsinxx3=sin3xx3×1cosx(cosx+1)\dfrac{\tan x - \sin x }{x^3} = \dfrac{\sin^3 x }{x^3} \times \dfrac{1}{\cos x (\cos x+1)}

Can you see where you might have gone wrong?


I get that but what's wrong with what i done to it? same result isn't it?

Posted from TSR Mobile
(edited 10 years ago)
Reply 4
No, your first equation is incorrect. Your first step actually only holds true if

limx0tan(x)x3<\displaystyle\lim_{x \rightarrow 0} \dfrac{\tan(x)}{x^3} < \infty

and

limx0sin(x)x3<\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x^3} < \infty

The problem is, the neither are true so your solution is invalid (which is why you've got the wrong answer). What you've done is turned it into an indeterminate \infty - \infty form, which does not necessarily go to 0.

You should have, from algebra of limits, that if:

f(x)Af(x) \rightarrow A, g(x)Bg(x) \rightarrow B as xpx \rightarrow p then limxp(αf(x)+βg(x))=αlimxpf(x)+βlimxpg(x)=αA+βB\displaystyle\lim_{x \rightarrow p} (\alpha f(x) + \beta g(x)) = \alpha \displaystyle\lim_{x \rightarrow p} f(x) + \beta \displaystyle\lim_{x \rightarrow p} g(x) = \alpha A + \beta B where α,βC\alpha, \beta \in \mathbb{C} but, more importantly, A,BA, B are finite (and the converse of this does not hold)

Just to reassure you that algebra of limits doesn't work like this, a good counterexample is to look the following

limx0(1+x)1x\displaystyle\lim_{x \rightarrow 0} \dfrac{(1+x) - 1}{x}

Well, we can easily work this out directly as we have

limx0(1+x)1x=limx0(1+xx1x)=limx0(xx)=1\displaystyle\lim_{x \rightarrow 0} \dfrac{(1+x) - 1}{x} = \displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1+x}{x} - \dfrac{1}{x} \right) = \displaystyle\lim_{x \rightarrow 0} \left( \dfrac{x}{x} \right) = 1

However, if we do what you have done, we get this:

limx0(1+x)1x=limx0(1+xx1x)=limx0(1x1x)=0\displaystyle\lim_{x \rightarrow 0} \dfrac{(1+x) - 1}{x} = \displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1+x}{x} - \dfrac{1}{x} \right) = \displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1}{x} - \dfrac{1}{x} \right) = 0

Which essentially comes down to saying you can't take the limit of one part before others.
(edited 10 years ago)
Original post by cooldudeman
I get that but what's wrong with what i done to it? same result isn't it?

Posted from TSR Mobile


No, see Noble's post above.
Reply 6
Original post by Noble.
No, your first equation is incorrect. Your first step actually only holds true if

limx0tan(x)x3<\displaystyle\lim_{x \rightarrow 0} \dfrac{\tan(x)}{x^3} < \infty

and

limx0sin(x)x3<\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x^3} < \infty

The problem is, the neither are true so your solution is invalid (which is why you've got the wrong answer). What you've done is turned it into an indeterminate \infty - \infty form, which does not necessarily go to 0.

You should have, from algebra of limits, that if:

f(x)Af(x) \rightarrow A, g(x)Bg(x) \rightarrow B as xpx \rightarrow p then limxp(αf(x)+βg(x))=αlimxpf(x)+βlimxpg(x)=αA+βB\displaystyle\lim_{x \rightarrow p} (\alpha f(x) + \beta g(x)) = \alpha \displaystyle\lim_{x \rightarrow p} f(x) + \beta \displaystyle\lim_{x \rightarrow p} g(x) = \alpha A + \beta B where α,βC\alpha, \beta \in \mathbb{C} but, more importantly, A,BA, B are finite (and the converse of this does not hold)

Just to reassure you that algebra of limits doesn't work like this, a good counterexample is to look the following

limx0(1+x)1x\displaystyle\lim_{x \rightarrow 0} \dfrac{(1+x) - 1}{x}

Well, we can easily work this out directly as we have

limx0(1+x)1x=limx0(1+xx1x)=limx0(xx)=1\displaystyle\lim_{x \rightarrow 0} \dfrac{(1+x) - 1}{x} = \displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1+x}{x} - \dfrac{1}{x} \right) = \displaystyle\lim_{x \rightarrow 0} \left( \dfrac{x}{x} \right) = 1

However, if we do what you have done, we get this:

limx0(1+x)1x=limx0(1+xx1x)=limx0(1x1x)=0\displaystyle\lim_{x \rightarrow 0} \dfrac{(1+x) - 1}{x} = \displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1+x}{x} - \dfrac{1}{x} \right) = \displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1}{x} - \dfrac{1}{x} \right) = 0

Which essentially comes down to saying you can't take the limit of one part before others.


ah OK I get ya. I got 1/2 after doing it again. so I should always turn it into a form that will avoid anything related to infinity?

Posted from TSR Mobile
Reply 7
Original post by cooldudeman
ah OK I get ya. I got 1/2 after doing it again. so I should always turn it into a form that will avoid anything related to infinity?

Posted from TSR Mobile


You certainly don't want to be splitting a limit up in such a way that one of them tends to infinity, because you won't get the right answer (and if you do, it's purely out of luck). You really have to try and do all the manipulation within the limit and try and get it into a form you can take the limit of in one step. You can do it in multiple steps, but you have to be very careful and make sure you know all the algebra of limit 'rules' (and ensure they are finite, most importantly).

Quick Reply