Limit that involves the angle addition and subtraction of tangent

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wolly

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[math]\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}[/math]
I know that

[math]\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}[/math]
and

[math]\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}[/math]
Do You multiply each fraction with this (1-tan(a)tan(x))(1+tan(a)tan(x))?
I expanded the tangents and now I'm stuck.
 
In my textbook the answer is [math]\frac{4sin(2a)}{cos^3(a)}[/math]
 
The answer that I got after calculating is
[math]2tan(a)+tan^2(a)[/math]
 
wolly said:
[math]\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}[/math]
I know that

[math]\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}[/math]
and

[math]\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}[/math]
Do You multiply each fraction with this (1-tan(a)tan(x))(1+tan(a)tan(x))?
I expanded the tangents and now I'm stuck.
Click to expand...
The two answers you show are not equivalent (as seen by graphing them); so you'll need to show us your actual work in order for us to see where you went wrong. Everything you said so far sounds fine.
 
wolly said:
[math]\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}[/math]
I know that

[math]\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}[/math]
and

[math]\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}[/math]
Do You multiply each fraction with this (1-tan(a)tan(x))(1+tan(a)tan(x))?
I expanded the tangents and now I'm stuck.
Click to expand...
[imath]\displaystyle \lim_{x\rightarrow 0}\frac{\tan(a + x) + \tan(a - x) - 2\tan a}{x^2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{\tan(x + a) - \tan(x - a) - 2\tan a}{x^2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{\sec^2(x + a) - \sec^2(x - a)}{2x}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{2[ \ \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) \ ]}{2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0} \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) = \ ?[/imath]
 
mario99 said:
[imath]\displaystyle \lim_{x\rightarrow 0}\frac{\tan(a + x) + \tan(a - x) - 2\tan a}{x^2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{\tan(x + a) - \tan(x - a) - 2\tan a}{x^2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{\sec^2(x + a) - \sec^2(x - a)}{2x}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0}\frac{2[ \ \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) \ ]}{2}[/imath]


[imath]= \displaystyle \lim_{x\rightarrow 0} \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) = \ ?[/imath]
Click to expand...
If I don't use L'hopital rule, I get this:

[imath]\displaystyle \lim_{x\rightarrow 0}\frac{\tan(a + x) + \tan(a - x) - 2\tan a}{x^2}[/imath]


[imath]\displaystyle = \lim_{x\rightarrow 0}\frac{2\tan^3 a \tan^2 x + 2 \tan^2 x \tan a}{x^2(1 - \tan a \tan x)(1 + \tan a \tan x)}[/imath]


[imath]\displaystyle = \lim_{x\rightarrow 0}\frac{2\tan^3 a \tan^2 x + 2 \tan^2 x \tan a}{x^2(1 - \tan^2 a \tan^2 x)}[/imath]

The next step is straightforward. Can you continue from here?

Note: I used the angles sum definition of [imath]\tan(a + x) \ \text{and} \ \tan(a - x)[/imath] in post #1.
 
And the answer that I get îs
[math]2tan^3(a)+2tan(a)[/math]
This îs not
[math]\frac{4sin(2a)}{cos^3(a)}[/math]
 
The correct answer is:

[imath]\displaystyle 2\tan^3 a + 2\tan a = 2\tan a(\tan^2 a + 1) = 2\tan a \sec^2 a = \frac{2\sin a}{\cos a}\times\frac{1}{\cos^2 a} = \frac{2\sin a}{\cos^3 a}[/imath]

L'hopital rule gives the same answer, so it is confirmed that it is the correct answer.

wolly said:
[math]\frac{4sin(2a)}{cos^3(a)}[/math]
Click to expand...
If this is the book answer, then there is a typo I think. If this is your answer, you should revise your steps to arrive to my steps in post #9.
 
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