# How to solve this limit without using L'Hopital

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.

• Try to use MathJax to type mathematical expressions. Images for such simples expressions are often frowned upon at MSE. Feb 3 at 14:09
• Try the mean value theorem: Ifr$F$ and $G$ differentiable over $[\alpha,\beta]$, then there is $\xi\in(\alpha,\beta)$ such that $G'(\xi)(F(\beta)-F(\alpha))=F'(\xi)(G(\beta)-G(\alpha))$. Feb 3 at 14:12
• Have you tried to use sum identities? Feb 3 at 14:25
• Using the identity $\tan a-\tan b=\tan(a-b) (1+\tan a\tan b)$ one can show that the numerator of the expression under limit is $\tan h\tan 2h\tan (a+h) (1+\tan a\tan (a+2h))$ and now the job is not that difficult. Feb 4 at 3:58
• @user: yes they could have used the same identity to deal with $\tan(a+2h)-\tan a$ at the end instead of using the derivative. But then everyone has the right to think differently :) Feb 4 at 11:06

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

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

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 $$\lim_{x\to0}{\sin x\over x}=1$$

• Avoiding derivatives and more advanced methods, this seems the more effective approach to me among the ones here given. Maybe you should mention which identities you are referring. for the last step it seems clear you are using standard limits.
– user
Feb 4 at 10:09
• gotcha ill put an edit Feb 4 at 10:44

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}

• Your second approach is the simplest out of all the answers here. +1 Feb 4 at 3:52

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,

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

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

• Why we can split the limit sign and change the order with h? Feb 4 at 0:39
• @EricLey, because the limit actually exists. Observe that how simple the calculation has become after $$\cos a-\cos(a+2h)?$$ Feb 4 at 4:05
• Your equation at the end is totally non-rigorous. A proper approach is to assume the existence of $f''(a)$ and apply L'Hospital's Rule. Feb 4 at 11:01