Homework Help: Can someone check this limit please

mtayab1994 · Feb 22, 2012

1. The problem statement, all variables and given/known data

find the limit of:

[tex]\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}[/tex]

3. The attempt at a solution

[tex]\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}=\frac{1-1-\frac{x^{2}}{2}}{tan(x)}=0[/tex]

lanedance · Feb 22, 2012

if you're talking about using taylor expansions, then you should do the same thing for tan(x) to show its behaviour in the limit, and your expnasion is not correct due to the 2 in cos(2x)

however as this is 0/0 indeterminate I think you should use L'Hopital (which is in essence a similar thing)

kushan · Feb 22, 2012

you are doing right

SammyS · Feb 22, 2012

mtayab1994 said: ↑

1. The problem statement, all variables and given/known data

find the limit of:

[tex]\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}[/tex]

3. The attempt at a solution

[tex]\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}=\frac{1-1-\frac{x^{2}}{2}}{tan(x)}=0[/tex]

That should be
[itex]\displaystyle \frac{1-\cos(2x)}{\tan(x)}=\frac{1-\left(1-\frac{(2x)^{2}}{2}+\dots\right)}{\tan(x)}=\frac{1-1+\frac{(2x)^{2}}{2}-\dots}{\tan(x)}[/itex]

You can do this without using L'Hôpital's rule or Taylor series.

Start with cos(2x) = 1 - 2sin²(x) and 1/tan(x) = cos(x)/sin(x) .

Mark44 · Feb 22, 2012

mtayab1994 said: ↑

1. The problem statement, all variables and given/known data

find the limit of:

[tex]\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}[/tex]

3. The attempt at a solution

[tex]\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}=\frac{1-1-\frac{x^{2}}{2}}{tan(x)}=0[/tex]

The lim symbol should remain until you actually take the limit. In the expression in the middle, above, the denominator is 0, so you can't conclude that the limit is zero.

kushan said: ↑

you are doing right

See above.

Mark44 · Feb 22, 2012

SammyS said: ↑

That should be
[itex]\displaystyle \frac{1-\cos(2x)}{\tan(x)}=\frac{1-\left(1-\frac{(2x)^{2}}{2}+\dots\right)}{\tan(x)}=\frac{1-1+\frac{(2x)^{2}}{2}-\dots}{\tan(x)}[/itex]

You can do this without using L'Hôpital's rule or Taylor series.

Start with cos(2x) = 1 - 2sin²(x) and 1/tan(x) = cos(x)/sin(x) .

This is definitely the way to go.

mtayab1994 · Feb 22, 2012

Mark44 said: ↑

This is definitely the way to go.

after doing that i got:

[tex]\lim_{x\rightarrow0}\frac{-sin^{2}(x)*cos(x)}{sin(x)}=-sin(x)*cos(x)=-sin(0)*cos(0)=0*1=0[/tex]

Is that correct?

Mark44 · Feb 22, 2012

mtayab1994 said: ↑

after doing that i got:

[tex]\lim_{x\rightarrow0}\frac{-sin^{2}(x)*cos(x)}{sin(x)}=-sin(x)*cos(x)=-sin(0)*cos(0)=0*1=0[/tex]

Is that correct?

No. How did you get from (1 - cos(2x))/tan(x) to (-sin²(x)cos(x))/sin(x)? Show us what you did to get that.

Also, the lim symbol needs to appear until you actually take the limit. In other words, as long as x is still in any expression, you haven't takent the limit.

mtayab1994 · Feb 22, 2012

Mark44 said: ↑

No. How did you get from (1 - cos(2x))/tan(x) to (-sin²(x)cos(x))/sin(x)? Show us what you did to get that.

Also, the lim symbol needs to appear until you actually take the limit. In other words, as long as x is still in any expression, you haven't takent the limit.

Well, i replaces 1-cos(2x) with 1-1-sin^2(x) and the bottom from tan(x) to sin(x)/cos(x)

Then i removed the denominator by 1/tan(x).

Mark44 · Feb 22, 2012

mtayab1994 said: ↑

Well, i replaces 1-cos(2x) with 1-1-sin^2(x) and the bottom from tan(x) to sin(x)/cos(x)

You have two mistakes. That should be 1 - (1 - 2sin²(x)).

mtayab1994 said: ↑

Then i removed the denominator by 1/tan(x).

mtayab1994 · Feb 22, 2012

alright and for the denominator should i do tanx=sinx/cosx?

mtayab1994 · Feb 22, 2012

Ok i got (2sin^2(x)*cos(x))/(sin(x))

then i canceled out with the sine and i was left with 2sinx*cosx which then equals 0 is that correct?

Mark44 · Feb 22, 2012

mtayab1994 said: ↑

alright and for the denominator should i do tanx=sinx/cosx?

Yes.

mtayab1994 said: ↑

Ok i got (2sin^2(x)*cos(x))/(sin(x))

then i canceled out with the sine and i was left with 2sinx*cosx which then equals 0 is that correct?

2sin(x)cos(x) doesn't equal 0, but the limit as x approaches 0 of this expression is 0.

In other words,
2sin(x)cos(x) ≠ 0, but [itex]\lim_{x \to 0} 2sin(x)cos(x) = 0[/itex]. There's a difference here.

mtayab1994 · Feb 22, 2012

Mark44 said: ↑

Yes.
2sin(x)cos(x) doesn't equal 0, but the limit as x approaches 0 of this expression is 0.

In other words,
2sin(x)cos(x) ≠ 0, but [itex]\lim_{x \to 0} 2sin(x)cos(x) = 0[/itex]. There's a difference here.

Yea sorry i didn't write, that but is what i did correct?

Mark44 · Feb 22, 2012

mtayab1994 said:

Yea sorry i didn't right that but is what i did correct?

The value of the limit is 0, but what you wrote is not correct.

mtayab1994 said:

i was left with 2sinx*cosx which then equals 0

Again, 2sinx * cosx ≠ 0, in general.

mtayab1994 · Feb 22, 2012

Mark44 said: ↑

The value of the limit is 0, but what you wrote is not correct.

Again, 2sinx * cosx ≠ 0, in general.

I don't get it what's wrong with that. Isn't the limit as x approaches 0 of 2sinx*cosx=0?

Mark44 · Feb 22, 2012

mtayab1994 said: ↑

I don't get it what's wrong with that. Isn't the limit as x approaches 0 of 2sinx*cosx=0?

Yes, but what you said didn't include the word "limit," so isn't true.

True:
[tex]\lim_{x \to 0}~2sin(x)cos(x) = 0[/tex]
False:
2sin(x)cos(x) = 0

I'm am trying to get you to distinguish between the value of an expression and the value of the limit of that expression.

Do you see the difference now?

mtayab1994 · Feb 22, 2012

Mark44 said: ↑

Yes, but what you said didn't include the word "limit," so isn't true.

True:
[tex]\lim_{x \to 0}~2sin(x)cos(x) = 0[/tex]
False:
2sin(x)cos(x) = 0

I'm am trying to get you to distinguish between the value of an expression and the value of the limit of that expression.

Do you see the difference now?

Yes thank you I appreciate it .

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Homework Help: Can someone check this limit please

mtayab1994

lanedance

kushan

SammyS

Mark44

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Mark44

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mtayab1994

Mark44

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mtayab1994

Mark44

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mtayab1994

mtayab1994

Mark44

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mtayab1994

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mtayab1994

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