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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.

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Just an update, I forgot to specify a bit more: I have to find a solution using trigonometric relations/identities, without using more advanced methods. – Mr.Anderson Jul 22 '13 at 7:06
up vote 6 down vote accepted

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 $0$.

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Thank you for the quick response T.Bongers. The problem is that I have been asked to come up with a solution to evaluate this limit using only trigonometric relations, and not more advanced calculus "tools". This is a great solution though, and I really appreciate you finding the time to answer me. Have a good one! – Mr.Anderson Jul 22 '13 at 7:04
Thanks, glad you liked it. I've added another technique that is a direct computation only involving trig rules. Cheers! – user61527 Jul 22 '13 at 7:13
Thank you for a very thorough explanation. Perfect conclusion along the lines of André Nicolas and Git Gud's great ideas. I thank you all for your time. Have a good one! – Mr.Anderson Jul 22 '13 at 7:19
@GitGud Fixed it, thanks. I always forget to do that... – user61527 Jul 22 '13 at 7:25
@T.Bongers No problem. Another tip: \$\left( \frac{a}{b}\right)\$ yields $\left( \dfrac{ a}{ b}\right)$ which looks better than $(\dfrac{ a}{ b})$. The \left \right thing works for all delimiters. – Git Gud Jul 22 '13 at 7:27

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?

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Thank you very much for your help. This was great to get me on track to solve this problem, and I actually found the solution continuing where you left off. I really appreciate your time in helping me out. Have a good one. – Mr.Anderson Jul 22 '13 at 7:21
You are welcome. The "tricks" I used come up moderately often. – André Nicolas Jul 22 '13 at 7:25

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,

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

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Thank you for taking the time to help me out. Have a great day. – Mr.Anderson Jul 22 '13 at 7:24
@Mr.Anderson: You are welcome. – Mhenni Benghorbal Jul 22 '13 at 7:26

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