baldymcgee6
  • baldymcgee6
Find the value of this limit:
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
baldymcgee6
  • baldymcgee6
http://gyazo.com/4ad8e332ba4e68323866548f1b84f5bb
hartnn
  • hartnn
write x^2-1 as (x+1)(x-1)
baldymcgee6
  • baldymcgee6
Okay, did that, I suspecting something will cancel somewhere?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
there is an established proof that limit z-->0 (sin(z))/z is 1 put z = x+1 z-->0 => x-->-1 write denominator as (x-1)*(x+1) limit will be (1/2)*1 which is 0.5
baldymcgee6
  • baldymcgee6
@kulprit the limit is going to -1 not 0, doesn't that matter?
anonymous
  • anonymous
Should be -1/2, I think...
anonymous
  • anonymous
limit z-->0 (sin(z))/z is 1 and so limit x-->-1 (sin(x+1))/(x+1) is 1 just substitute k = x+1 x-->-1 ,so x+1-->0 => k-->0
anonymous
  • anonymous
yeah sorry @Jemurray3 limit is -0.5 disregard that
baldymcgee6
  • baldymcgee6
I understand how the lim x->0 sin(x)/x = 1 but I don't understna dhow we can use this when this limit is going to -1, sorry for being difficult :)
baldymcgee6
  • baldymcgee6
@Jemurray3, can you help me understand how we can use this trig limit even when x approaches -1 and not 0?
anonymous
  • anonymous
you have lim (x->-1) sin(x+1)/(x+1)
anonymous
  • anonymous
let u = x+1. As x -> -1, u -> 0, so the above is the same as lim (u -> 0 ) sin(u)/u
anonymous
  • anonymous
should be 1/(x-1) I believe but yes, that's the idea.
baldymcgee6
  • baldymcgee6
\[\frac{ 1 }{ x-1 } \lim_{x \rightarrow -1} \frac{ \sin(x+1) }{ x+1 }\]
baldymcgee6
  • baldymcgee6
how can we say thay u ->0?
anonymous
  • anonymous
because (x+1) goes to zero as x goes to -1.
baldymcgee6
  • baldymcgee6
ooohhh gotcha
baldymcgee6
  • baldymcgee6
what do we do with the 1/(x-1)
anonymous
  • anonymous
Nothing, that's perfectly continuous as x approaches -1 so that just becomes -1/2.
baldymcgee6
  • baldymcgee6
is that where we get the -0.5?
baldymcgee6
  • baldymcgee6
oh.. k good. thanks so much
anonymous
  • anonymous
Oh, I wasn't paying attention... that should be on the right side of the limit. so it should be \[ \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x^2-1} = \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x+1}\cdot \frac{1}{x-1} \] \[ = \frac{-1}{2} \]
baldymcgee6
  • baldymcgee6
oh right, because it's not a constant right?
anonymous
  • anonymous
right.
baldymcgee6
  • baldymcgee6
thanks, you rock.

Looking for something else?

Not the answer you are looking for? Search for more explanations.