Limit that involves the angle addition and subtraction of tangent

[wolly]

$$\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}$$
I know that

$$\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}$$
and

$$\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}$$
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.

 

[wolly]

In my textbook the answer is $$\frac{4sin(2a)}{cos^3(a)}$$

 

[wolly]

The answer that I got after calculating is
$$2tan(a)+tan^2(a)$$

 

[Dr.Peterson]

$$\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}$$
I know that

$$\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}$$
and

$$\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}$$
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.

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]

 

[wolly]

I forgot to place the limits

 

[wolly]

Tg îs how we note tangent in my country
Tg=tan

 

[mario99]

$$\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}$$
I know that

$$\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}$$
and

$$\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}$$
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.

$\displaystyle \lim_{x\rightarrow 0}\frac{\tan(a + x) + \tan(a - x) - 2\tan a}{x^2}$


$= \displaystyle \lim_{x\rightarrow 0}\frac{\tan(x + a) - \tan(x - a) - 2\tan a}{x^2}$


$= \displaystyle \lim_{x\rightarrow 0}\frac{\sec^2(x + a) - \sec^2(x - a)}{2x}$


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


$= \displaystyle \lim_{x\rightarrow 0} \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) = \ ?$

 

[mario99]

$\displaystyle \lim_{x\rightarrow 0}\frac{\tan(a + x) + \tan(a - x) - 2\tan a}{x^2}$


$= \displaystyle \lim_{x\rightarrow 0}\frac{\tan(x + a) - \tan(x - a) - 2\tan a}{x^2}$


$= \displaystyle \lim_{x\rightarrow 0}\frac{\sec^2(x + a) - \sec^2(x - a)}{2x}$


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


$= \displaystyle \lim_{x\rightarrow 0} \sec^2(x + a)\tan(x + a) - \sec^2(x - a)\tan(x - a) = \ ?$

If I don't use L'hopital rule, I get this:

$\displaystyle \lim_{x\rightarrow 0}\frac{\tan(a + x) + \tan(a - x) - 2\tan a}{x^2}$


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


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

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

Note: I used the angles sum definition of $\tan(a + x) \ \text{and} \ \tan(a - x)$ in post #1.

 

[wolly]

And the answer that I get îs
$$2tan^3(a)+2tan(a)$$
This îs not
$$\frac{4sin(2a)}{cos^3(a)}$$

 

[mario99]

The correct answer is:

$\displaystyle 2\tan^3 a + 2\tan a = 2\tan a(\tan^2 a + 1) = 2\tan a \sec^2 a = \frac{2\sin a}{\cos a}\times\frac{1}{\cos^2 a} = \frac{2\sin a}{\cos^3 a}$

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

$$\frac{4sin(2a)}{cos^3(a)}$$

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.