calculus limit q Watch

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

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cooldudeman
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actually I think I got it.

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Indeterminate
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(Original post by cooldudeman)
actually I think I got it.

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Note that

\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?
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cooldudeman
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(Original post by Indeterminate)
Note that

\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?

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Noble.
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No, your first equation is incorrect. Your first step actually only holds true if

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

and

\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) \rightarrow A, g(x) \rightarrow B as x \rightarrow p then \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 \alpha, \beta \in \mathbb{C} but, more importantly, A, 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

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

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

\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:

\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.
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Indeterminate
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(Original post by cooldudeman)
I get that but what's wrong with what i done to it? same result isn't it?

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

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

and

\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) \rightarrow A, g(x) \rightarrow B as x \rightarrow p then \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 \alpha, \beta \in \mathbb{C} but, more importantly, A, 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

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

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

\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:

\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?

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Noble.
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(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?

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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).
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