- #1
- 584
- 0
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]
That should beHomework 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]
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
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) .
after doing that i got:This is definitely the way to go.
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.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?
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.
You have two mistakes. That should be 1 - (1 - 2sin^{2}(x)).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).
Yes.alright and for the denominator should i do tanx=sinx/cosx?
2sin(x)cos(x) doesn't equal 0, but the limit as x approaches 0 of this expression is 0.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?
Yea sorry i didn't write, that but is what i did correct?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.
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
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.
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?
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 the limit of that expression.
Do you see the difference now?