Why is this way of doing $\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$

Question

I did it this way :

$$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$ now as $x \to \frac{\pi}{2}$, $\tan(2*x) \to \tan(\pi)$ and $x - \frac{\pi}{2} \to 0$
As $\tan(\pi) = 0$

$$\therefore \tan(2*x) \to 0$$

$\therefore$ we can write $$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$ As $$\lim_{x \to 0} \frac{\tan(x)}{x} = 1$$

Which is incorrect answer.

My second attempt :

$$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$ $$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{\cos(2*x)(x - \frac{\pi}{2})}$$ $$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{(x - \frac{\pi}{2})} * \lim_{x \to \frac{\pi}{2}} (\frac{1}{\cos(2*x)})$$ $$-\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{(x - \frac{\pi}{2})}$$ $$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{\color{red}{(\frac{\pi}{2} - x)}}$$ $$\lim_{x \to \frac{\pi}{2}} \frac{2 * \sin(2*x)}{\pi - 2 *x}$$ $$2 * \lim_{x \to \frac{\pi}{2}} \frac{\sin(\pi - 2*x)}{\pi - 2 *x} $$ Because, $$\sin(x) = \sin(\pi - x)$$ $$\therefore 2 * \lim_{x \to \frac{\pi}{2}} \frac{\sin(\pi - 2*x)}{\pi - 2 *x} = 2 $$

Which is correct and makes sense.

Now i did this question one more way

$$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$ $$\lim_{x \to \frac{\pi}{2}} \frac{2 * \tan(2*x)}{2x - \pi}$$ $$2 * \lim_{x \to \frac{\pi}{2}} -\frac{\tan(\pi - 2*x)}{2*x - \pi}$$ $$2 * \lim_{x \to \frac{\pi}{2}} \frac{\tan(\pi - 2*x)}{\pi-2*x}$$ Now we observe that now as $x \to \frac{\pi}{2}$, $\tan(\pi -2*x) \to 0$ and $\pi - 2*x \to 0$ Thus we can write

$$2 * \lim_{x \to \frac{\pi}{2}} \frac{\tan(\pi - 2*x)}{\pi-2*x} = 2 * \lim_{x \to 0} \frac{\tan(x)}{x} = 2$$

Which is also correct, Can anyone please tell me what is incorrect with 1 approach ?

@ritwiksinha try evaluating $\lim_{x \to 0} \sin(nx)/x$ the way you did in your first approach. — Nikunj, Jun 06 '16 at 06:16
@Nikunj Oh yes. thanks that was a great way of explaining things. — , Jun 06 '16 at 06:22

score 1 · Accepted Answer · answered Jun 06 '16 at 06:02

1

$$\dfrac{\tan2x}{x-\dfrac\pi2}=\dfrac{\tan2\left(x-\dfrac\pi2\right)}{x-\dfrac\pi2}=2\cdot\dfrac{\tan(2x-\pi)}{2x-\pi}$$

Actually we have $$\lim_{(h)\to0}\dfrac{\tan(h)}{(h)}=\lim_{h\to}\dfrac{\sin h}h\cdot\dfrac1{\lim_{h\to0}\cos h}=1$$

Observe that all the $h$ in parenthesis must be same.

answered Jun 06 '16 at 06:02

lab bhattacharjee

274,582

But why should it be like that ? Because both of them is approaching the same value. – Jun 06 '16 at 06:05
@ritwiksinha, See http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – lab bhattacharjee Jun 06 '16 at 06:07
@ritwiksinha You could also take a look at http://math.stackexchange.com/questions/1795207/extended-lim-x-rightarrow-0-frac-sinxx-1-limit-law – ApprenticeOfMathematics Jun 06 '16 at 06:57
@ApprenticeOfMathematics But that doesn't answer why both numerator and denominator has to be equal of it to work. i.e $\sin(\frac{"smallThing"}{"Bigthing"})$, if both big thing and small thing is tending towards zero why can we put $x$ in both places. – Jun 06 '16 at 10:26
@ritwiksinha Do you mean something like $\lim_{\ h\to0}\dfrac{\sin \ (small\ thing)}{(bigger \ thing)}$ where both the "small thing" and "bigger thing" is tending to zero? Can you give me an example? – ApprenticeOfMathematics Jun 06 '16 at 14:59
@ApprenticeOfMathematics $\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$ – Jun 06 '16 at 16:04
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@ritwiksinha: You have to have both the variables in numerator and denominator to be same and tend to $0$. This is because the standard result $(\sin x)/x \to 1$ as $x \to 0$ has same variable $x$ in both numerator and denominator. The result is not like $(\sin x)/y \to 1$ when $x, y $ both tend to $0$ (and this is false if $x \neq y$). – Paramanand Singh Jun 10 '16 at 07:07

Why is this way of doing $\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$

1 Answers1