Calculate trig limit of type $\frac{0}{0}$ without L'Hopital

Question

I am trying to figure out the solution to this Limit without using L'Hopital.

$$ \lim \limits_{x \to \pi} \frac {(\tan (4x))^2 } {(x - \pi )^2} $$

Any help would be greatly appreciated.

Answer 1

One could evaluate the Taylor series expansion of the numerator around $x = \pi$; then division shows that

$$\frac{\tan(4x)^2}{(x - \pi)^2} = 16 + O((x - \pi)^2)$$

Informally, the second piece will tend to zero as $x$ tends to $\pi$. A formal convergence argument could then be made to justify that the limit is, in fact, 16.

Edit: Another way to do it without using a series to only use trigonometric relations as follows:

$$\tan(4x)^2 = \frac{\sin(4x)^2}{\cos(4x)^2} = \frac{\sin(4x - 4\pi)^2}{\cos(4x - 4\pi)^2}$$

where we have used that $\sin$ is $2\pi$-periodic. Thus, we can rewrite the original limit as

$$\lim_{x\to\pi} \frac{\tan(4x)^2}{(x - \pi)^2} = \frac{1}{\cos(4x - 4\pi)^2}(\frac{\sin(4x - 4\pi)}{(x - \pi)})^2$$

Clearly, the term involving cosine tends to 1. On the other hand, let $u = (x - \pi)/4$; then we may rewrite the limit as

$$\lim_{u \to 0} (\frac{\sin(u)}{u/4})^2 = 16 \lim_{u \to 0} \frac{\sin(u)}{u} = 16$$

recalling that $\sin(u)/u$ tends to 1 as $u$ tends to $0$.

Answer 2

Here is a start

$$\lim_{x\to \pi} \frac{\tan(4x)^2}{(x-\pi)^2}=\lim_{y\to 0} \frac{\sin^2(4(y+\pi))}{y^2}\frac{1}{\cos^2(y+\pi)}=\lim_{y\to 0} \frac{\sin^2{4y}}{y^2}=\dots. $$

Note that,

$$ \sin(4(y+\pi)) = \sin(4y). $$

Answer 3

Not necessary, but I would let $x-\pi=t$. Note that $\tan(4x)=\tan(4t+4\pi)=\tan(4t)$. We want the limit of $$\frac{\tan(4t)}{t}\cdot \frac{\tan(4t)}{t}$$ as $t\to 0$.

Note that $$\frac{\tan(4t)}{t}=\frac{4}{\cos(4t)}\frac{\sin(4t)}{4t}.$$

Now can you finish?