Question

lim x->pie/4 (1-tanx)/(sinx-cosx)

Answer 1

\[\lim_{x\to\pi}\frac{1-\tan x}{\sin x-\cos x}\]
step 1) replace tan by sin and cos

Answer 2

ok
then wat do I do

Answer 3

then things cancel themselves up :)

Answer 4

so I have (cosx-sinx/cosx)/sinx-cosx

Answer 5

use the equation editor below to enter your equation. otherwise, I cannot understand

Answer 6

so then I just cross cancel ?

Answer 7

ok one min plz

Answer 8

\[\left[ cosx-sinx/cosx \right]\div \left[ sinx-cosx \right]\]

Answer 9

so now I just cross cancel?

Answer 10

\[\lim_{x\to\pi}\Large \frac{-\cancel{(\sin x-\cos x)}\over\cos x}{\cancel{\sin x-\cos x}}\]

Answer 11

\[=\lim_{x\to\pi}{-1\over\cos x}\]

Answer 12

then I just plug in \[\pi/4\] for cosx

Answer 13

I didn't know I could do that within a limit

Answer 14

thank u very much :)

Answer 15

you can divide if and only if it is not zero.

by definition, as \(x\to\pi/4\), \(x\ne\pi/4\)
and hence, \(\sin x\ne\cos x\implies\sin x-\cos x\ne 0\)
hence, they can be cancelled. :o

Answer 16

oh ok thank u :)

Answer 17

yw.

Answer 18

uh one question if u don't mind

Answer 19

shoot it

Answer 20

um I don't know how u got the negative bcz \[\left( 1-tanx/ sinx-cosx \right)\]
\[\left[ 1-sinx / cosx \right] / (sinx-cosx)\]
them cross cancel and I get

Answer 21

\[\left[ 1/cosx \right]\]

Answer 22

so I don't know where the negative comes from

Answer 23

\[1-{\sin x\over\cos x}={\cos x-\sin x\over \cos x}\]
look the factors you are cancelling!!

Answer 24

\[(a-b)=-(b-a)\]

Answer 25

yes then I cancel that out with \[\left[ 1divsinx-cosx \right]\]

Answer 26

take you time with the equation editor :)
atleast I will know what you are saying!

Answer 27

sorry its my first time

Answer 28

that is ok.
Welcome to Open Study :D

Answer 29

\[\frac{ cosx-sinx }{ cosx } \times \frac{ 1 }{ sinx-cosx }\] then I cross cancel so where is the negative comeing from ...thanx for the the
welcome

Answer 30

you cannot cancel as yet. they are not same...
one says \((a-b)\)
the other says \((b-a)\)

Answer 31

ok yes I understand that

Answer 32

then, can you make them look same? yes by taking out the "negative" sign

Answer 33

\[\frac{-(\sin x-\cos x)}{\cos x}\times{1\over \sin x-\cos x}\]

Answer 34

if I take out the negative sign from \[\left( cosx-sinx \right)\] then wouldn't it be \[-\left( sinx+cosx \right)\]

Answer 35

ooopsss wait

Answer 36

I mean it would be \[-\left( cosx+sinx \right)\]

Answer 37

\[a-b=-(-a)-(b)=-(-a+b)=-(b-a)\]

Answer 38

you have to flip all the signs. positive become negative and negatives become positives

Answer 39

oh my gosh I got it im so sorry I confused my self more than necessary hehe

Answer 40

:)
happens to the best of us. This is an indication that you learnt something.

Answer 41

yup most definitely, thank u for ur help :)

Answer 42

you are welcome.