Question

Limit question, to be posted below.

Answer 1

\[\lim_{x \rightarrow \pi} \frac{ \tan(tanx) }{ x-\pi }\]

Answer 2

apply L'Hospital Rule

diferrentiate separately the numerator and the denominator

Answer 3

not allowed to use that rule

Answer 4

\[\tan= \sin \pi/\cos\]

Answer 5

@drpruitt how does that help us? and how did you get there?

Answer 6

Are you allowed to used any rule but L'H Rule???

Answer 7

Like what?

Answer 8

@UnkleRhaukus, @satellite73, @Hero any ideas?

Answer 9

not without l'hopital no
what else could you possible use?

Answer 10

trig limits

Answer 11

this question was just on my test, there has to be a way to do it without L' Hospital rule

Answer 12

for a composition of functions? if you have a non l'hopital i would love to see it

Answer 13

@lgbasallote

Answer 14

I remember doing this in calc. class actually...
I think you use the fact that lim x->pi (tanx) = x-pi ....(the line)

Answer 15

but how were we supposed to know that?!

Answer 16

sometimes useless facts come in handy

Answer 17

hmm, there's got to be a better way.

Answer 18

I say Pinching theorem

Answer 19

and what would that look like?

Answer 20

lim x->0 sinx/x = 1

Answer 21

You're gonna have to help me out a little more :)

Answer 22

But have you learned derivatives, at least?

Answer 23

yes

Answer 24

Might I suggest you "sneakily" use l'Hopital?
for instance, since
\[\tan(\tan \pi) = 0 \]anyway, you could subtract it from the numerator with no "horrid consequences"
\[\lim_{x \rightarrow \pi}\frac{\tan(\tan x)-\tan(\tan \pi)}{(x - \pi)-(\pi - \pi)}\]
And then divide both the numerator and denominator by (x - pi)
\[\huge \lim_{x \rightarrow \pi}\frac{\frac{\tan(\tan x)-\tan(\tan \pi)}{x - \pi}}{\frac{(x-\pi)-(\pi -\pi)}{x-\pi}}\]

Answer 25

tan(tanx)/x-pi
=sin(tanx)/cos(tanx) * 1/(x-pi)
=sin(tanx)/cos(tanx) * tanx/tanx * 1/(x-pi)
=sin(tanx)/tanx * tanx/cos(tanx) * 1/(x-pi)
= 1 * tanx/cos(tanx) * 1/(x-pi)

this is as far as i can get, i cant get the cos(tanx) to disappear

Answer 26

But yeah, since the limits of both the numerator and the denominator exist
since
tan(tan x) and (x - pi) are both differentiable, then that expression is just equal to
\[\huge \frac{\lim_{x \rightarrow \pi}\frac{\tan(\tan x)-\tan(\tan \pi)}{x - \pi}}{\lim_{x \rightarrow \pi}\frac{(x-\pi)-(\pi -\pi)}{x-\pi}}\]
or,
if
\[f(x) = \tan(\tan x) \ and \ g(x) = x - \pi\]
It is then equal to
\[\huge \frac{f'(\pi)}{g'(\pi)}\]
Not exactly l'Hôpital, I grant, but it works...

Answer 27

Sorry, I overdid it, all you had to do after all was subtract
tan (tan pi) from the numerator, since the denominator was already x - pi:
\[ \lim_{x \rightarrow \pi}\frac{\tan(\tan x) - \tan(\tan \pi)}{x-\pi}=\left[ \frac{d}{dx}\tan(\tan x) \right]_{x=\pi}\]

Answer 28

Thank you so much @terenzreignz

Answer 29

No problem :)

Answer 30

For anyone interested: http://screencast.com/t/6AH3ecG5kLh
@terenzreignz, @satellite73, @Algebraic!, @Minh_Minh, @drpruitt

Answer 31

it's blank... post SS?

Answer 32

SS?

Answer 33

If that link doesn't work, here is an upload.

Answer 34

So they did use Pinching theorem for this one but how can (x-pi)=z and x=z also?