Epsilon - delta proof $\lim_{x \to \frac{\pi}{4}} \tan(x)=1$

Question

I need to prove the limit using the definition of limit $$\lim_{x\to c}f(x)=L \leftrightarrow \forall \epsilon >0 \hspace{0.5 cm} \exists \delta >0 : 0<|x-c|<\delta \rightarrow |f(x)-L|<\epsilon $$

My attempt $$|\tan(x)-1|<\epsilon\\ -\epsilon <\tan(x)-1 < \epsilon\\ \tan^{-1}(-\epsilon+1)<x<\tan^{-1}(\epsilon +1)\\ \tan^{-1}(-\epsilon+1)- \frac{\pi}{4}<x-\frac{\pi}{4}<\tan^{-1}(\epsilon +1)-\frac{\pi}{4}$$ I don't know if with this I can give an adequate value for $\delta$. Any advice on how to solve it would be very helpful, thank you.

Answer 1

Here is a more explicit way to do it, without using arctan of epsilon.

Since $\tan(x)$ has an asymptote at $x=\frac{\pi}{2}$, we should stipulate at the outset a maximum value of delta which is less than the distance from the point we are working at and the asymptote (this distance is $\frac{\pi}{4}$), for instance $\delta<\frac{\pi}{8}$. So in what follows we will consider only $x \in (\frac{\pi}{4}-\frac{\pi}{8},\frac{\pi}{4}+\frac{\pi}{8})=(\frac{\pi}{8},\frac{3\pi}{8})$. Now, if we suppose that $\epsilon$ is arbitrary and $0<|x-\frac{\pi}{4}|<\delta$ holds, we have

$$ |\tan(x)-1| = \left|\frac{\sin(x)}{\cos(x)}-\frac{\sin \left(\frac{\pi}{4} \right)}{\cos \left(\frac{\pi}{4} \right)} \right| = \left|\frac{\sin(x)\cos \left(\frac{\pi}{4} \right)-\sin \left(\frac{\pi}{4} \right) \cos(x)}{\cos(x)\cos \left(\frac{\pi}{4} \right)} \right|=\frac{|\sin\left(x-\frac{\pi}{4} \right)|}{|\cos(x)|\cos \left(\frac{\pi}{4} \right)}$$ Now in the numerator we can use the inequality $|\sin \left(x-\frac{\pi}{4} \right)| \leq |x-\frac{\pi}{4}|$ while to bound the denominator we notice that cosine is decreasing on the interval $(\frac{\pi}{8},\frac{3\pi}{8})$ and so $\frac{1}{|\cos(x)|} <\frac{1}{\cos (\frac{3\pi}{8})}$. These two inequalities allow us to write $$ \frac{|\sin\left(x-\frac{\pi}{4} \right)|}{|\cos(x)|\cos \left(\frac{\pi}{4} \right)} < \frac{|x-\frac{\pi}{4}|}{\cos \left(\frac{3\pi}{8} \right) \cos \left(\frac{\pi}{4} \right)}<\frac{\delta}{\cos \left(\frac{3\pi}{8} \right) \cos \left(\frac{\pi}{4} \right)} \leq \epsilon $$ If we chose $\delta = \min \{\frac{\pi}{8}, \cos \left(\frac{3\pi}{8} \right)\cos \left(\frac{\pi}{4} \right) \epsilon \}$.

Answer 2

Alternative approach:

$\underline{\textbf{Preliminary Results }}$

Result-1
$\displaystyle \lim_{x \to 0} ~\tan(x) = 0.$

Proof:
Attempting to stay within the spirit of the problem,
since the sine function is continuous, I will assume that :

• $\displaystyle \lim_{x \to 0} \sin(x) = \sin(0) = 0.$

Edit
If you are given that $~ \displaystyle \lim_{x \to 0}\frac{\sin(x)}{x} = 1, ~$
then the above assumption is a consequence of that.

For $~\epsilon > 0,~$ set $~ \displaystyle \epsilon_1 = \frac{\epsilon}{\sqrt{2}}.$
$\exists ~\delta_1 > 0~$ such that $~|x| \leq \delta_1 \implies |\sin(x)| < \epsilon_1$.
Also, $~\displaystyle |x| \leq (\pi/4) \implies \frac{1}{\sqrt{2}} \leq \cos(x) \leq 1.$

For $\epsilon > 0,~$ set $~\delta = \min(\delta_1, \pi/4).$
Then $~\displaystyle |x| \leq \delta \implies$

• $\displaystyle |\sin(x)| < \frac{\epsilon}{\sqrt{2}}$

• $\displaystyle \frac{1}{\sqrt{2}} \leq |\cos(x)| \leq 1 \implies 1 \leq \frac{1}{\cos(x)} \leq \sqrt{2}.$

Therefore, $~\displaystyle |x| \leq \delta = \min(\delta_1, \pi/4) \implies $

$\displaystyle |\tan(x)| = \left|\frac{\sin(x)}{\cos(x)}\right| < \frac{\epsilon}{\sqrt{2}} \times \sqrt{2} = \epsilon.$

Result-2
$\displaystyle \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}.$

Proof:
See wikipedia trig angle sum identities.

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$\underline{\textbf{Main Problem}}$

Given $~\epsilon > 0,~$ set $\epsilon_2 = \min[(1/2), \epsilon/5].$

Invoke Result-1.
Then, there exists a $~\delta_2 ~$ such that
$-\delta_2 < y < \delta_2 \implies |\tan(y)| < \epsilon_2.$

Suppose $-\delta_2 < [x - \pi/4] < \delta_2.$
Set $y = [x - \pi/4] \implies$

• $x = (y + \pi/4) \implies \tan(x) = \tan(y + \pi/4)$
• $-\delta_2 < y < \delta_2 \implies -\epsilon_2 < \tan(y) < \epsilon_2.$

Invoke Result-2.

Then

$$\tan(x) = \tan(y + \pi/4) = \frac{\tan(y) + 1}{1 - \tan(y)}.\tag1 $$

In (1) above, let $N$ denote the numerator and let $D$ denote the denominator.

Then $1 - \epsilon_2 < N < 1 + \epsilon_2$
and $1 - \epsilon_2 < D < 1 + \epsilon_2.$

Therefore

$$\frac{1 - \epsilon_2}{1 + \epsilon_2} < \frac{\tan(y) + 1}{1 - \tan(y)} < \frac{1 + \epsilon_2}{1 - \epsilon_2}. \tag2 $$

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However, because $~\epsilon_2 = \min[(1/2), \epsilon/5],~$ you have that

• $\displaystyle 1 - \epsilon < 1 - 2\epsilon_2 < 1 - \frac{2\epsilon_2}{1 + \epsilon_2} = \frac{1 - \epsilon_2}{1 + \epsilon_2}.$

• $\displaystyle \frac{1 + \epsilon_2}{1 - \epsilon_2} = 1 + \frac{2\epsilon_2}{1 - \epsilon_2} \leq 1 + \frac{2\epsilon_2}{(1/2)} = 1 + 4\epsilon_2 < 1 + \epsilon.$

Therefore $\tan(x) = \tan(y + \pi/4)$ and

$$1 - \epsilon < \tan(y + \pi/4) < 1 + \epsilon.$$