Question

need help on lim -->1 of sin(pi*x)/(x-1)

Answer 1

use substitution x-1 = u, x= u+1
limit u -> 0
and see what happens

Answer 2

hum.. let me see.

Answer 3

can't you use L'Hospital's rule,@experimentX since you have the indeterminant form 0/0?

Answer 4

L'Hoptal's is not alowed.

Answer 5

yes you can ... but not recommended if you can do algebraically.

best substitution is to use x-1 = u/pi

Answer 6

:( ok.

Answer 7

I get stucked on lim u-->0 of sin(u+1)/u

Answer 8

no that is not correct form

Answer 9

sin x/ x = 1 how to apply this?

Answer 10

replace the x in sin ( pi*x) with x = u+1

Answer 11

wait

Answer 12

I got lim u-->0 -sin(u*pi)/u

Answer 13

now what?

Answer 14

(-sin(u)/u)*sin(pi)?

Answer 15

-1*sin(pi)?

Answer 16

=0

Answer 17

use another substitution \( \pi u = v \)

Answer 18

ok, let me try...

Answer 19

lim v->0 - sin(v)/v making v=pi*u
= -1?

Answer 20

change the lower u to v/pi

Answer 21

you forgot to add pi to the top

Answer 22

sorry, I'm lost, where i did the mistake?

Answer 23

I forgot to add pi? where?

Answer 24

\[ \frac{-\sin (\pi u)}{u} = ?\]

Answer 25

.sin(pi*u+pi) = sin(pi*u)cos(pi) + sin(pi)cos(pi*u)
sin (pi) = 0
and cos(pi) = -1
then -sin(pi*u)... where is the missing pi?

Answer 26

put u = v/pi

you get
-sin(pi*v/pi)/(v/pi) = - pi sin(v)/v = -pi

Answer 27

Nice. Thank you. But I need more exercicies like this. Can you indicate any link?

Answer 28

don't know ... if you have used this x-1 = u/pi, you would have gotten easy answer.

don't know ... try constructing few stuff yourself.

Answer 29

ok. Thank you again!

Answer 30

yw