Limit question- Fairly easy? (1 Viewer)

BLIT2014 · Sep 5, 2015

Screen Shot 2015-09-05 at 5.55.08 pm.png

Thank you

InteGrand · Sep 5, 2015

BLIT2014 said:
View attachment 32365

Thank you

$$\frac{\cos2x -1}{x}=\frac{1-2\sin^2 x -1}{x}=\frac{-2\sin^2 x}{x}$.$

$\therefore \lim _{x\to 0}\frac{\cos2x -1}{x}=\lim_{x\to 0}\frac{-2\sin^2 x}{x}$

$=\lim_{x\to 0}\left(-2\sin x \times \frac{\sin x}{x} \right)$

$=\lim_{x\to 0}\left(-2\sin x\right) \times \lim_{x\to 0}\left(\frac{\sin x}{x} \right)$

$=0\times1$

$$=0$.$

InteGrand · Sep 5, 2015

The limit exists and is 0.

BLIT2014 · Sep 5, 2015

InteGrand said:
$$\frac{\cos2x -1}{x}=\frac{1-2\sin^2 x -1}{x}=\frac{-2\sin^2 x}{x}$.$

$\therefore \lim _{x\to 0}\frac{\cos2x -1}{x}=\lim_{x\to 0}\frac{-2\sin^2 x}{x}$

$=\lim_{x\to 0}\left(-2\sin x \times \frac{\sin x}{x} \right)$

$=\lim_{x\to 0}\left(-2\sin x\right) \times \lim_{x\to 0}\left(\frac{\sin x}{x} \right)$

$=0\times1$

$$=0$.$

Thank you, I appreciate it.

Oer · Sep 5, 2015

While I wouldn't recommend using it in HSC working out, using L'Hopital's rule could be useful to check your answer in questions like this, just to make sure it's right. Look it up if you haven't heard of it.

InteGrand · Sep 5, 2015

Oer said:
While I wouldn't recommend using it in HSC working out, using L'Hopital's rule could be useful to check your answer in questions like this, just to make sure it's right. Look it up if you haven't heard of it.

Also, use the calculator to check if your answer looks right.

Silly Sausage · Sep 5, 2015

L'hoptial's rule:
$\lim _{x\to c} \frac{f(x)}{g(x)} = \lim _{x\to c} \frac{f'(x)}{g'(x)}$

Silly Sausage · Sep 5, 2015

Silly Sausage said:
L'hoptial's rule:
$\lim _{x\to 0} \frac{f(x)}{g(x)} = \lim _{x\to 0} \frac{f'(x)}{g'(x)}$

So $\lim _{x\to 0} \frac{cos2x-1}{x}$
=
$\lim _{x\to 0} \frac{-2sin2x}{1}=0$

dan964 · Sep 5, 2015

Silly Sausage said:
So $\lim _{x\to 0} \frac{cos2x-1}{x}$
=
$\lim _{x\to 0} \frac{-2sin2x}{1}=0$

does L'Hospitale's rule is such that you stated, provided the original limit exists?

InteGrand · Sep 5, 2015

dan964 said:
isn't that provided the original limit exists?

It's actually provided the latter limit exists, which it does.

dan964 · Sep 5, 2015

thanks
couldn't remember which one, I had read about the rule before

psyc1011 · Sep 5, 2015

Silly Sausage said:
L'hoptial's rule:
$\lim _{x\to 0} \frac{f(x)}{g(x)} = \lim _{x\to 0} \frac{f'(x)}{g'(x)}$

I encourage OP (and others) to use L'Hopital's rule as last resort because if you begin with L'Hop, all you are testing is your memory and understanding a formula.

dan964 · Sep 5, 2015

agreed

Silly Sausage · Sep 5, 2015

There have been marker's comments on how people used this method in the HSC exams.

Sien · Sep 5, 2015

So if you use lhop. Rule do you still get the marks or nah

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InteGrand · Sep 5, 2015

Sien said:
So if you use lhop. Rule do you still get the marks or nah

Sent from my D6503 using Tapatalk

If you do use it, make sure you know (and state in your answer) the conditions that are required to hold for it to be applicable.

Drongoski · Sep 5, 2015

$\lim _ {x \to 0} \frac {cos 2x - 1}{x} = \lim _ {x \to 0} \frac {-2sin ^2 x}{x} = -2 \lim _ {x \to 0} \frac {sin x}{x} \cdot sin x \\ \\ = -2 \lim _ {x \to 0}\frac {sin x}{x} \cdot \lim _ {x \to 0} sin x = -2 \times 1 \times 0 = 0$

No need for L'Hopital's; not in HSC syllabus.

Edit

Didn't see InteGrand's similar approach, at the beginning.

RealiseNothing · Sep 6, 2015

There are a few other nit-picky things such as the limit has to be indeterminate and g'(x) =/= 0 over the domain (not including the point the limit is going to itself)

Limit question- Fairly easy? (1 Viewer)

BLIT2014

The pessimistic optimist.

InteGrand

Well-Known Member

InteGrand

Well-Known Member

BLIT2014

The pessimistic optimist.

Oer

Started from the fundus

InteGrand

Well-Known Member

Silly Sausage

Well-Known Member

Silly Sausage

Well-Known Member

dan964

what

InteGrand

Well-Known Member

dan964

what

psyc1011

#truth

dan964

what

Silly Sausage

Well-Known Member

Sien

将来: NEET

InteGrand

Well-Known Member

Drongoski

Well-Known Member

RealiseNothing

what is that?It is Cowpea

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