- #1

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## Homework Statement

find the limit of:

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

## 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]

- Thread starter mtayab1994
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- #1

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find the limit of:

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

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

- #2

lanedance

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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)

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

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- #3

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you are doing right

- #4

SammyS

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That should be## Homework Statement

find the limit of:

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

## 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]

[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 - 2sin

- #5

Mark44

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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.## Homework Statement

find the limit of:

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

## 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]

See above.you are doing right

- #6

Mark44

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This is definitely the way to go.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 - 2sin^{2}(x) and 1/tan(x) = cos(x)/sin(x) .

- #7

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after doing that i got:This is definitely the way to go.

[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?

- #8

Mark44

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No. How did you get from (1 - cos(2x))/tan(x) to (-sinafter 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?

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.

- #9

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Well, i replaces 1-cos(2x) with 1-1-sin^2(x) and the bottom from tan(x) to sin(x)/cos(x)No. How did you get from (1 - cos(2x))/tan(x) to (-sin^{2}(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.

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

- #10

Mark44

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

- #11

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

- #12

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

- #13

Mark44

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

2sin(x)cos(x) doesn't equal 0, but the

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

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.

- #14

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

2sin(x)cos(x) doesn't equal 0, but thelimitas 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.

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- #15

Mark44

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The value of the limit is 0, but what you wrote is not correct.mtayab1994 said:Yea sorry i didn't right that but is what i did correct?

Again, 2sinx * cosx ≠ 0, in general.mtayab1994 said:i was left with 2sinx*cosx which then equals 0

- #16

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I don't get it what's wrong with that. Isn't the limit as x approaches 0 of 2sinx*cosx=0?The value of the limit is 0, but what you wrote is not correct.

Again, 2sinx * cosx ≠ 0, in general.

- #17

Mark44

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Yes, but what you said didn't include the word "limit," so isn't true.I don't get it what's wrong with that. Isn't the limit as x approaches 0 of 2sinx*cosx=0?

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

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

Do you see the difference now?

- #18

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Yes thank you I appreciate it .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 thelimitof that expression.

Do you see the difference now?

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