Question

Limit problem.

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \sin( a+2x)-2\sin(a+x)+\sin(a)}{ x^2 }\]

Answer 2

So using \[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\] isn't enough here, right?

Answer 3

I wonder what happens if you try to expand the sin parts using the sum rule

Answer 4

Haha, do you really wonder or is that a hint?

Answer 5

it is really a wondering thing

Answer 6

and we can't use l'hosptal right?

Answer 7

Nah, this is in the chapter juuuust before they introduce L'hopital's th'm.

Answer 8

why is it ever time we get a limit problem it comes with "without l'hopital"
annoying is what that is

Answer 9

I think expanding it by sum rule will work
it will end up being really long though

Answer 10

I don't like it
there has to be a shorter way

Answer 11

=( I think it's the only way w/o L'hopital's th'm

Answer 12

Taylor polynomial gives a quick solution

Answer 13

not a tool in my belt ><

Answer 14

\[\sin(a+2x)+\sin(a)=2\sin(\frac{2a+2x}{2})\cos(\frac{2a-2x}{2}) \\ \sin(a+2x)+\sin(a)=2\sin(a+x)\cos(a-x)\]
So we could write it as \[\lim_{x \rightarrow 0}\frac{2 \sin(a+x)\cos(a-x)-2\sin(a+x)}{x^2}\]
we can now do some factoring...

Answer 15

I don't know if that is a dead end yet
haven't work it though

Answer 16

but it looks kinda nice

Answer 17

oops I made a mistake above

Answer 18

\[\sin(a+2x)+\sin(a)=2\sin(\frac{2a+2x}{2})\cos(\frac{2x}{2}) \\ \sin(a+2x)+\sin(a)=2\sin(a+x)\cos(x) \]

Answer 19

@KinzaN do you know where to go from there

Answer 20

like I used the identity \[\sin(x)+\sin(y)=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})\]

Answer 21

I will give you one more hint
this is what you should have after applying that identity

\[\lim_{x \rightarrow 0}\frac{2\sin(a+x)\cos(x)-2\sin(a+x)}{x^2} \\ \lim_{x \rightarrow 0}\frac{2\sin(a+x)(\cos(x)-1)}{x^2}\]

now multiply top and bottom by (cos(x)+1)

ok now I won't say anything else :p

Answer 22

Ahh thank you, I'll stare at it until it makes sense

Answer 23

lol

Answer 24

the identity didn't make sense?

Answer 25

or what I'm asking you to do next?

Answer 26

you probably evaluated this limit before:
\[\lim_{x \rightarrow 0}\frac{\cos(x)-1}{x} \text{ or } \lim_{x \rightarrow 0}\frac{1-\cos(x)}{x}\]
and you evaluated both of these by multiply the top and bottom by cos(x)+1

Answer 27

we are using that same kinda deal above you will definitely be using the limit
as x approaches 0 sin(x)/x approaches 1

Answer 28

\[\lim_{x \rightarrow 0}\frac{2\sin(a+x)\cos(x)-2\sin(a+x)}{x^2} \\ \lim_{x \rightarrow 0}\frac{2\sin(a+x)(\cos^2(x)-1)}{x^2(\cos(x)+1)} \]
recall cos^2(x)-1=-sin^2(x)
and what do you know about limit as x approaches 0 of sin^2(x)/x^2?

Answer 29

Ah, okay that I understand--you x by the reciprocal to be able to use the sinx/x limit. I'm confused about the first thing that you worked through though

Answer 30

so the identity that I applied

Answer 31

how you rewrote it ^^ b/c if that's an identity I've never seen it before

Answer 32

oo embarrassing haha is it just an identity

Answer 33

\[\sin(a)+\sin(b)=2 \sin(\frac{a+b}{2})\cos(\frac{a-b}{2})\]

there are so many identities lol
I don't remember them all honestly
I googled identities so I could look at a possible list of identities that could be used here to make things easier

Answer 34

http://mathforum.org/dr.math/faq/formulas/faq.trig.html

Answer 35

Think I could reach the same point without it? Using just the basic cos or sin(a+b) and sin(2a)

Answer 36

messier but it's probably what I should practice since that identity is beyond me lol