# advanced limits exercise with trigonometry

Evaluate $$\lim\limits_{h\to0} \frac{\tan(a+2h)-2\tan(a+h)+\tan a}{h}.$$

I have already tried to expand $$\tan (a+2h)$$ and $$\tan(a+h)$$ but it did not lead me anywhere.

As an alternative, using trigonometric exapansion we can write:

$$\frac{\tan(a+2h)-2\tan(a+h)+\tan a}{h}= \\ =\frac{\tan(a+2h)-\tan(a+h)}{h}-\frac{\tan(a+h)-\tan a}{h}$$ and by expansion for the first term $$\frac{\tan(a+2h)-\tan(a+h)}{h}=\frac{\tan(a+h)+\tan h-\tan(a+h)(1-\tan(a+h)\tan h)}{h(1-\tan(a+h)\tan h)}= \\=\frac{\tan h+\tan^2(a+h)\tan h}{h(1-\tan(a+h)\tan h)}=\frac{\tan h}{h}\frac{1+\tan^2(a+h)}{1-\tan (a+h)\tan h}\to 1\cdot (1+\tan^2 a)=\frac 1 {\cos^2 a}$$

and since

$$\frac{\tan(a+h)-\tan a}{h}\to \tan'a=\frac 1 {\cos^2 a}$$

the result follows.

Let's generalize!

Consider a differentiable map $$f$$. We want to evaluate

$$\lim\limits_{h \to 0} \frac{f(a+2h)-2 f(a+h) +f(a)}{h}$$

You have

$$\frac{f(a+2h)-2f(a+h)+f (a)}{h} = \frac{f(a+2h)-f(a)}{h} - \frac{f(a+h)-f( a)}{h}- \frac{f(a+h)-f(a)}{h}$$

For $$h \to 0$$, the first term of the RHS converges to $$2f^\prime(a)$$, the second and the third one to $$f^\prime(a)$$. Hence the limit is equal to zero.

Take $$\tan$$ for $$f$$. The limit is equal to zero.

• Took me a minute, but you're right. Got hung up on the $2f'(a)$. – Chris Custer Sep 23 at 8:59

Apply L'Hospital Rule then

$$L=\lim_{h\rightarrow 0} \frac{\tan(a+2h)-2\tan(a+h)-\tan a}{h}=lim_{h \rightarrow 0} ~2 sec^2(a+2h)-2 sec^2(a+h)=0.$$

$$\tan(a+2h)-2tan(a+h)+\tan a=\tan(a+2h)-\tan(a+h)-(\tan(a+h)-\tan a)$$

$$\dfrac{\tan(a+2h)-2\tan(a+h)+\tan a}h=\dfrac{\tan(a+2h)-\tan(a+h)}h-\dfrac{\tan(a+h)-\tan a}h$$

Now

$$\lim_{n\to0}\dfrac{\tan(a+(n+1)h)-\tan(a+nh)}h$$

$$=\lim_{n\to0}\dfrac{\sin h}h\cdot\lim_{n\to0}\cos\dfrac{a+(n+1)h+a+nh}2$$

$$=?$$