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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.

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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.

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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.

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  • $\begingroup$ Took me a minute, but you're right. Got hung up on the $2f'(a)$. $\endgroup$ – Chris Custer Sep 23 at 8:59
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Apply L'Hospital Rule then

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

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

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

Now

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

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

$=?$

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