Question

find the limit of x approaches 1 in cotx/(cscx+1)
i know the answer should be around (1/2) but i dont know how to get it
@wio

Answer 1

Hmmm

Answer 2

I would start by multiplying top and bottom by the conjugate, which is \(\csc x - 1\)

Answer 3

this would get me ((cotx)(cscx+1))/((cscx-1)(cscx+1))

Answer 4

what would i do from there?

Answer 5

any luck solving the problem?

Answer 6

Expand the bottom

Answer 7

how would expanding the bottom help?

Answer 8

foil

Answer 9

so i get csc^2x-1, where do i go from there

Answer 10

There is a trig identity for \(csc\) and \(\cot\)

Answer 11

which trig identity would it be?

Answer 12

http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm

Answer 13

Wait, this is the limit as \(x\to 1\)?

Answer 14

yes

Answer 15

any luck?

Answer 16

Okay \[
\csc^2 x-1= \cot^2x
\]This is what I was trying to get you to realize.

Answer 17

i got that part, so i got \[1/\cot ^2x\]

Answer 18

where do i go from there?

Answer 19

Well the \(\cot\) cancel out, learing only one \(\cot\) at the bottom.

Answer 20

what would the limit be then? \[\lim_{x \rightarrow 1} 1/\cot x\]

Answer 21

It should become: \[
\lim_{x\to 1}\frac{\csc x -1}{\cot x}
\]

Answer 22

\[
\lim_{x\to 1}\frac{\csc x -1}{\cot x}=
\lim_{x\to 1} (\tan x \csc x -\tan x)
\]

Answer 23

\[
\lim_{x\to 1}\frac{\csc x -1}{\cot x}=
\lim_{x\to 1} (\tan x \csc x -\tan x)=
\lim_{x\to 1} (\sec x -\tan x)
\]

Answer 24

Now you just take the limit. That should be pretty straightforward.