Can someone check this limit please

[mtayab1994]

Homework Statement

find the limit of:

$$\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}$$

The Attempt at a Solution

$$\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}=\frac{1-1-\frac{x^{2}}{2}}{tan(x)}=0$$

 

[lanedance]

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]

you are doing right

 

[SammyS]

Homework Statement

find the limit of:

$$\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}$$

The Attempt at a Solution

$$\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}=\frac{1-1-\frac{x^{2}}{2}}{tan(x)}=0$$

That should be

$\displaystyle \frac{1-\cos(2x)}{\tan(x)}=\frac{1-\left(1-\frac{(2x)^{2}}{2}+\dots\right)}{\tan(x)}=\frac{1-1+\frac{(2x)^{2}}{2}-\dots}{\tan(x)}$​

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]

Homework Statement

find the limit of:

$$\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}$$

The Attempt at a Solution

$$\lim_{x\rightarrow0}\frac{1-cos(2x)}{tan(x)}=\frac{1-1-\frac{x^{2}}{2}}{tan(x)}=0$$

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.

you are doing right

See above.

 

[Mark44]

That should be

$\displaystyle \frac{1-\cos(2x)}{\tan(x)}=\frac{1-\left(1-\frac{(2x)^{2}}{2}+\dots\right)}{\tan(x)}=\frac{1-1+\frac{(2x)^{2}}{2}-\dots}{\tan(x)}$​

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]

This is definitely the way to go.

after doing that i got:

$$\lim_{x\rightarrow0}\frac{-sin^{2}(x)*cos(x)}{sin(x)}=-sin(x)*cos(x)=-sin(0)*cos(0)=0*1=0$$

Is that correct?

 

[Mark44]

after doing that i got:

$$\lim_{x\rightarrow0}\frac{-sin^{2}(x)*cos(x)}{sin(x)}=-sin(x)*cos(x)=-sin(0)*cos(0)=0*1=0$$

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]

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]

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)).

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

 

[mtayab1994]

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

 

[mtayab1994]

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]

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

Yes.

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 $\lim_{x \to 0} 2sin(x)cos(x) = 0$. There's a difference here.

 

[mtayab1994]

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 $\lim_{x \to 0} 2sin(x)cos(x) = 0$. There's a difference here.

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

 

[Mark44]

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.

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

Again, 2sinx * cosx ≠ 0, in general.

 

[mtayab1994]

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]

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:
$$\lim_{x \to 0}~2sin(x)cos(x) = 0$$
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]

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

True:
$$\lim_{x \to 0}~2sin(x)cos(x) = 0$$
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 .