Question

limit as x->pi/2 of [tanx - 1/(x-pi/2)]

Answer 1

so i had this on a quiz today and i said it was undefined

Answer 2

looks like the limit will not exist. either that or you could say \(-\infty\)

Answer 3

yay! so i'm right?

Answer 4

i would have said that as well
so i bet you got it right

Answer 5

it was a quiz on l'hospital's rule

Answer 6

well you cannot use l'hopital here unless you change the form, because it is not \(\frac{0}{0}\)

Answer 7

and i was talking to someone who was saying that he first changed tanx to sinx/cosx then cross multiplied to make them have a common denominator, then was going to try to get a form of 0/0, but he didn't finish.

Answer 8

so i wasn't sure how to go about doing that or if that would work

Answer 9

i think that is too much work, but maybe you could do something like that
\[\frac{\sin(x)}{\cos(x)}-1=\frac{\sin(x)-\cos(x)}{\cos(x)}\]nah too much work

Answer 10

k cool.

Answer 11

so i guess my follow up question is, if i split the limit into lim tanx + lim (1/(x-pi/2)), then say they're both undefined because they approach infinity and negative infinity depending on the direction you're approaching pi/2 from, does that necessarily imply that the original limit is also undefined?

Answer 12

I think \(\infty-\infty\) is indeterminate, so you can't stop there

Answer 13

If the problem is
\[\lim_{x \rightarrow \frac{\pi}{2}}\tan(x)+\frac{1}{x-\frac{\pi}{2}}=0\]
It takes multiple steps to get to this answer.

First, it is not in the form 0/0 . we could do
\[\frac{\sin(x)}{\cos(x)} + \frac{1}{x-\frac{\pi}{2}}\]
\[ \frac{\sin(x)(x-\frac{\pi}{2})+\cos(x) } {\cos(x)(x-\frac{\pi}{2})}\]
which is now in 0/0 form.

the derivative of the numerator gives us
\[ \sin(x) +x \cos(x) -\frac{\pi}{2}\cos(x)-\sin(x) = x \cos(x) -\frac{\pi}{2}\cos(x)\]

the derivative of the denominator is
\[ -x\sin(x)+\cos(x)+\frac{\pi}{2}\sin(x) = \cos(x)+(\frac{\pi}{2}-x)\sin(x)\]
we again have 0/0

However, one more set of differentiations will give us \( \frac{0}{-2}=0\)