How to solve this limit without using L'Hopital

Question

How to solve this?

$$\lim_{h\rightarrow0}\frac{\tan(a+2h)-2\tan(a+h)+\tan a}{h^2}$$

I have solved it by L'Hopital but wasn't able to do it with other method.

One method i tried writing $\tan$ as $\sin/\cos$ which ended with some more worse equation.

Answer 1

By a simple but long way, we can use that

$$\tan(a+2h)=\frac{\tan a+\tan 2h}{1-\tan a\tan 2h}, \;\;\;\tan(a+h)=\frac{\tan a+\tan h}{1-\tan a\tan h}$$

and then

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

$$\require{cancel}\small=\frac{\cancel{\tan a}+\tan 2h-\cancel{\tan^2 a\tan h}-\tan a\tan h\tan 2h-\cancel{2\tan a}-2\tan h+2\tan^2 a \tan 2h+2\tan a \tan 2h \tan h}{h^2(1-\tan a\tan 2h)(1-\tan a\tan h)}\ldots$$

$$\ldots\small\frac{\cancel{\tan a}-\tan^2 a\tan 2h-\tan^2 a\tan h+\tan^3 a\tan 2h\tan h}{h^2(1-\tan a\tan 2h)(1-\tan a\tan h)}=$$

$$\small=\frac{(1+\tan^2 a)\tan 2h+(-2-\tan^2a)\tan h}{h^2(1-\tan a\tan 2h)(1-\tan a\tan h)}+ \frac{(\tan a+\tan^3 a)\tan h \tan 2h}{h^2(1-\tan a\tan 2h)(1-\tan a\tan h)}\to 2\frac{\tan a}{\cos^2 a}$$

indeed by tangent half-angle identity and standard limits

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

and therefore

$$\small \frac{(1+\tan^2 a)\tan 2h+(-2-\tan^2a)\tan h}{h^2(1-\tan a\tan 2h)(1-\tan a\tan h)}=\frac{2(1+\tan^2 a)\tan^3 h}{h^2(1-\tan a\tan 2h)(1-\tan a\tan h)(1-\tan^2 h)} \to 0$$

and by standard limits

$$\small \frac{(\tan a+\tan^3 a)\tan h \tan 2h}{h^2(1-\tan a\tan 2h)(1-\tan a\tan h)} \to 2(\tan a+\tan^3 a)=2\tan a(1+\tan^2 a)=2\frac{\tan a}{\cos^2 a}$$

Answer 2

By mean value theorem for function $F(t)=\tan(a+h+t)-\tan(a+h)-[\tan(a+t)-\tan a]$, we have $${\tan(a+2h)-\tan(a+h)-[\tan(a+h)-\tan a]=[\tan'(a+h+\theta)-\tan'(a+\theta)]h}$$ for some $|\theta|\leq |h|$ Use mean value theorem again, $$\tan(a+2h)-2\tan(a+h)+\tan a=\tan''(a+\lambda)h^2$$ for some $|\lambda|\leq 2|h|$. so the result is $\tan''(a)$.

You can also use diference identities for $\tan$: $\tan x-\tan y=\tan(x-y)(1+\tan x\tan y)$ \begin{align}&\tan(a+2h)-2\tan(a+h)+\tan a\\=&\tan h(1+\tan(a+2h)\tan(a+h))-\tan h(1+\tan(a+h)\tan(a))\\=&\tan h\tan (a+h)(\tan (a+2h)-\tan(a))\end{align} so \begin{align}&\lim_{h\to 0}\frac{\tan h\tan (a+h)(\tan (a+2h)-\tan(a))}{h^2}\\ =&\lim_{h\to 0}\frac{\tan h}{h}\cdot \tan (a+h)\cdot\frac{\tan (a+2h)-\tan(a)}{h}\\ =&\tan a \cdot 2\tan 'a\\=&2 \sec^2a \tan a\end{align}

Answer 3

Hint:

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

Now $\cos a-\cos(a+2h)=?$

Finally use $\lim_{h\to0}\dfrac{\sin h}h=1$

More generally,

$$\lim_{h\to0}\dfrac{\lim_{h\to0}\dfrac{f(a+2h)-f(a+h)}h-\lim_{h\to0}\dfrac{f(a+h)-f(a)}h}h$$

$$=\lim_{h\to0}\dfrac{f'(a+h)-f'(a)}h=f''(a)$$

Answer 4

If you are comfortable with differentiation limits this is just $\tan''(a)=2\sec^2(a)\tan(a)$

But a solution can also be done with pure trig this would ofc mirror the process of finding the derivative of tan

$$ \begin{align} &\lim_{h\to0}{(\tan(a+2h)-\tan(a+h))-(\tan(a+h)-\tan(a))\over h^2}\\ \ \\ =&\lim_{h\to0}{{\sin(h)\over\cos(a+2h)\cos(a+h)}-{\sin(h)\over\cos(a+h)\cos(a)}\over h^2}\\ \ \\ =&\lim_{h\to0}{\sin(h){\cos(a)-\cos(a+2h)\over\cos(a+2h)\cos(a+h)\cos(a)}\over h^2}\\ \ \\ =&\lim_{h\to0}{2\sin(h)\sin(a+h)\sin(h)\over h^2\cos(a+2h)\cos(a+h)\cos(a)}=2\sec^2(a)\tan(a) \end{align} $$

Incase any of the identities are not familiar line 1 is $$ \tan A - \tan B = {\sin A\over\cos A}-{\sin B\over\cos B}={\sin A\cos B-\sin B\cos A\over\cos A\cos B}={\sin (A-B) \over \cos A\cos B} $$

And a standard limit in the last step $$ \lim_{x\to0}{\sin x\over x}=1 $$