Question

Need help solving the limit.

Answer 1

\[\Large \lim_{x \rightarrow π/4 }~\frac{\sin(x+\frac{1}{4}π)-1}{x-\frac{1}{4}π}\]

\[HINT:~~~~~x+π/4=x-π/4+π/2\]

Answer 2

this is correct problem.

Answer 3

hmm i dont know what hint is trying to suggest.

do you know L'hopital's rule?

Answer 4

Yes, derive top and bottom, but it is only when it is approaching ∞ or 0.

Answer 5

I imagined using the hint to rewrite sine in terms of the angle addition rule. We might be aiming at getting a sin(u) / u limit.

Answer 6

The hint isn't a perfect square, and I am getting stuck b/c of this.

Answer 7

This is a begging of calculus, before (officially) doing derivatives

Answer 8

L'hopital's rule can be used regardless to what x approaches to . it only can be used when lim get you 0/0 or ∞/∞

Answer 9

Yeah? tnx greeky, but I need to do it without L'H'S because it is from Calculus 2 with I haven't yet learn. I am stuck on this prob, despite that I can do all others in the section.

Answer 10

LH is from Calculus 2? O_o I learned it in Calculus 1...

Answer 11

Well, I am not up to it yet

Answer 12

sorry, i don't really know another way, using hint. you can do it? @AccessDenied

Answer 13

I tried subbing 180 for π, but all I get on the top is

( √2/2 ) [ sin(x)-cos(x) ] -1

Answer 14

Access, please go slow on me ....

Answer 15

Actually not the sin u / u limit, but its friend may help:
http://www.proofwiki.org/wiki/Limit_of_(Cosine_(X)_-_1)_over_X
But we take sin (x + pi/4) = sin ((x - pi/4) + pi/2) and rewrite it using a trig identity for angle addition:
\( \sin (a + b) = \sin a \cos b + \sin b \cos a \)

Answer 16

Yes, I will get 1-cos if I derive the top, but I can't. This problem is before the derivatives in the book.

Answer 17

Saying derive top and bottom....

Answer 18

Not derivative, this is a trig identity.

Answer 19

but how am I supposed to translate this sine into cosine ?

Answer 20

I was going to suggest sine addition formula, but actually it might be easier if we know the straight relationship of sine and cosine by shift:
\( \sin \left(A + \pi/2 \right) = \cos A \)
Does that one look familiar?

Answer 21

aha that it.

Answer 22

Jesus I got disconnected.

Answer 23

\[\Large \lim_{x \rightarrow π/4}~\frac{\sin(x+\frac{1}{4}π)- 1 } { x-\frac{1}{4}π } \times \frac{x+ \frac{1}{4}π}{x+\frac{1}{4}π} \]

Answer 24

\[\Large \lim_{x \rightarrow π/4}~\frac{\sin(x+\frac{1}{4}π)- 1 } { x+\frac{1}{4}π } \times \frac{x+ \frac{1}{4}π}{x-\frac{1}{4}π} \]

Answer 25

\[\Large \lim_{x \rightarrow π/4}~\frac{- 1 } { x+\frac{1}{4}π } \times \frac{x+ \frac{1}{4}π}{x-\frac{1}{4}π} \]

Answer 26

This is how I am starting. I skipped the step of
(a-b)/c = a/c -b/c

Answer 27

\[\Large \lim_{x \rightarrow π/4}~\times \frac{(x+ \frac{1}{4}π)^{2}}{-1(x-\frac{1}{4}π)} \]

Answer 28

Idk, this is what I am getting so far -;(

Answer 29

no multiplication sign in the last thing ... ignore

Answer 30

I like the idea, but yeah that factor x - pi/4 is stuck in the denominator some way or another!
What I was suggesting starts out immediately with that hint.
\( \displaystyle \Large \lim_{x \rightarrow π/4 }~\frac{\sin(\color{blue}{x+\frac{1}{4}π})-1}{x-\frac{1}{4}π} \)
We rewrite that blue part as \(x - \pi/4 + \pi/2 \)
\( \displaystyle \Large \lim_{x \rightarrow π/4 }~\frac{\sin(\color{blue}{(x - \frac{1}{4} \pi) + \frac{1}{2} \pi})-1}{x-\frac{1}{4}π} \)
So that now we can rewrite sine into a shifted cosine as sin (A + pi/2) = cos A

Answer 31

where do you get sin(x+π/2) from? you forgot -π/4 I think.

Answer 32

sin(A + pi/2) = cos A is a general statement. This is always true, sine is just a cosine curve that is shifted over a bit!
Replace A with x - pi/4 and we still have a true statement; we can switch sin((x - pi/4) + pi/2) = cos(x - pi/4)

Answer 33

So the answer is zero, and it turns into the theorem you showed from wiki ?

Answer 34

Yes! x approaching pi/4 for cos(x - pi/4) - 1 / (x - pi/4) is the same as if u = x - pi/4, u is approaching 0 and we had cos u - 1 / u. :)

Answer 35

Yeah, thank you. I didn't know the "sin(a+π/2)=cos(a) " that was very very helpful !!!!!!!!!

Answer 36

No problem! It could have been done by using sine angle addition formula as well if you didn't know the sine - cosine relation, but it is more convenient with that identity. :D

Answer 37

Yes, definitely the theorem of sin(a+π/2)=cos(a) is better. You are great !!!