Question

lim x approaches pi (1-sin(x/2)/pi-x) cant use LHospitals rule ;/

Answer 1

tried substituting
y = pi-x
?

as x -> pi
x-pi ->0
y->0

Answer 2

0/0 ..

Answer 3

come on noone knows this?

Answer 4

what did you get for

sin(x/2)in terms of y ?

Answer 5

sin(pi-y/2) ??if x=pi-y

Answer 6

you mean
sin ((pi-y)/2)
?

which is
sin (pi/2 - y/2)

what does that simplify to ?

Answer 7

whaat?
sorry I am not that good in english? do you, by simplify, mean that there is some kind of trig equation that is egual to sin(pi/2-y/2) ?? because if you ment to input y->0 the result is again 0/0

Answer 8

yes, i mean there is a simplified trigonometric expression for

sin (pi/2-y/2)

Answer 9

sin (pi/2 - theta) = cos theta

Answer 10

ok let me try thanks

Answer 11

nope still 0/0

Answer 12

Something that might also be useful: \(\cos x=-\cos(x-\pi)\).

Answer 13

don't plug in 0 now...
unless you cancel out a common factor from numerator and denominator
(or using standard formulas like sin x/x =1)

Answer 14

what limit you have till now ?

Answer 15

lim y->0
cos (y/2)/y
right ?

Answer 16

[1-cos(y/2)]/y

Answer 17

tried all that and every supstitution there is..working on this for two days...
right

Answer 18

ok
then multiply and divide by 1+cos(y/2)

and try to simplify the numerator
(1-cos(y/2))(1+cos(y/2))

Answer 19

Try it a slightly different way (with same result):
\[\large\begin{align*}\lim_{x\to\pi}\frac{1-\sin\dfrac{x}{2}}{\pi-x}&=\lim_{x\to\pi}\frac{1-\sin\dfrac{x}{2}}{\pi-x}\cdot\frac{1+\sin\dfrac{x}{2}}{1+\sin\dfrac{x}{2}}\\\\
&=\lim_{x\to\pi}\frac{1-\sin^2\dfrac{x}{2}}{(\pi-x)\left(1+\sin\dfrac{x}{2}\right)}\\\\
&=\lim_{x\to\pi}\frac{1-\dfrac{1-\cos x}{2}}{(\pi-x)\left(1+\sin\dfrac{x}{2}\right)}\\\\
&=\frac{1}{2}\lim_{x\to\pi}\frac{1+\cos x}{(\pi-x)\left(1+\sin\dfrac{x}{2}\right)}\\\\
&=\frac{1}{4}\lim_{x\to\pi}\frac{1+\cos x}{\pi-x}\\\\
&=\frac{1}{4}\lim_{x\to\pi}\frac{1-\cos (x-\pi)}{\pi-x}\\\\
&=\frac{1}{4}\lim_{x\to\pi}\frac{1-\cos (\pi-x)}{\pi-x}\end{align*}\]
which strongly resembles a known limit,
\[\large\lim_{x\to c}\frac{1-\cos (x-c)}{x-c}=\color{red}{?}\]

Answer 20

0/0 again??

Answer 21

... There's no way to tell you what the next step is without telling you the answer outright.

Answer 22

\[\text{Do you know what }\lim_{x\to0}\frac{\sin x}{x}\text{ is?}\]

Answer 23

Hint: NOT 0/0

Answer 24

i know that the answer is zero and i know to get to it by LHospitals rule but i do not know how to get a zero without the rule

Answer 25

And you don't need L'Hopital's rule. You just need to recall a *known* limit and be able to apply it here.

Answer 26

in line 5, how did you get 1/2 out of (1+sin (x/2)) ?

Answer 27

i mean 2

Answer 28

I used the property that\[\lim_{x\to c}f(x)g(x)=\lim_{x\to c}f(x)\times\lim_{x\to c}g(x)\]

Answer 29

you could also recall the definition of derivative \[\lim_{x \rightarrow \pi}\frac{f(x)-f(\pi)}{x-\pi}=f'(x)|_{x=\pi} \\\text{ and you have something that looks similar } \\ \lim_{x \rightarrow \pi}\frac{\sin(\frac{x}{2})-1}{x-\pi}\]

Answer 30

cant figure this out.... ://study electrical engineering they said, it will be fun they said...

Answer 31

Which method do you prefer? Using algebra to rewrite the limit so that it simplifies to something you (should) know, or the definition of the derivative as per freckles's excellent suggestion? (The second is much more elegant in retrospect)

Answer 32

the first one please :D

Answer 33

love algebra :D

Answer 34

Alright, well if you can follow up to the last step, then you have a solid understanding of the algebraic manipulation involved here.

The very last step is calculus, or rather remembering things you would usually learn in a calc course.

Answer 35

we dont have that kind of subjects on our university only mathematichs which implies everything tigether, algebra, geometry, functions and so on..but I will try to find solution to final step.. thanks guys :)

Answer 36

You don't get offered a course in calculus, yet you have an engineering department? Suspicious.