Lim

[Guest]

[tex]\lim_{x \to 1} (x-1)tan(x\pi/2)[/tex]
How can i solve this without L'Hopital's Rule
Help!

[HallsofIvy]

So you want to do this the HARD WAY?

Okay, I would start by letting y= x-1. Then x= y+ 1 and, as x goes to 1, y goes to 0.
So the limit becomes [tex]\lim_{y\to 0}y tan((\pi/2)y+ \pi/2)[/tex].

Of course, [tex]tan(x)= \frac{sin(x)}{cos(x)}[/tex] so this limit is [tex]\lim_{y\to 0}y \frac{sin((\pi/2)y+ \pi/2)}{cos((\pi/2)y+\pi/2)}[/tex].

Now, [tex]sin(x+ \pi/2)= cos(x)[/tex] and [tex]cos(x+ \pi/2)= -sin(x)[/tex] so we can write that limit as
[tex]\lim_{y\to 0} y\frac{cos((\pi/2)y)}{-sin((\pi/2)y)}[/tex].

Let [tex]u= \left(\frac{\pi}{2}\right)y[/tex]. As y goes to 0, u also goes to 0 and we can write the limit as [tex]\lim_{u\to 0}\frac{2}{\pi}u\frac{cos(u)}{-sin(u)}= -\frac{2}{\pi}\lim_{u\to 0}\left(\frac{u}{sin(u)}\right)\left(cos(u)\right)= -\frac{2}{\pi}\left(\lim_{u\to 0}\frac{u}{sin(u)}\right)\left(\lim_{u\to 0} cos(u)\right)[/tex].

Now, I am sure you know that cos(x) is continuous for all x so that [tex]\lim_{u\to 0} cos(u)= cos(0)= 1[/tex].

What about [tex]\lim_{u\to 0} \frac{u}{sin(u)}[/tex]? (Perhaps you have seen [tex]\lim_{x\to 0}\frac{sin(x)}{x}[/tex].)