welshfella
  • welshfella
Need help with this one:- Find limit as x --> 0 of 2csc^2 x - (1/2) csc^2 (1/2) x.
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
welshfella
  • welshfella
both terms tend to infinity as x approaches zero so i converted them to the form f(x)/g(x) in order to apply l'hopitals but and got it to the form (0/0). Thats as far as i got.
ganeshie8
  • ganeshie8
i think we can try to simplify the expression using trig identities
ganeshie8
  • ganeshie8
\[\begin{align} 2\csc^2 x - \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1-\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align}\] we can take the limit now

Looking for something else?

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

More answers

welshfella
  • welshfella
yea
welshfella
  • welshfella
1/2
thomas5267
  • thomas5267
Not really awake. Use with caution. \[ \begin{align*} &\phantom{=}2\csc^2(x)-\frac{1}{2}\csc^2\left(\frac{1}{2}x\right)\\ &=\frac{2}{\sin^2(x)}-\frac{1}{2\sin^2\left(\frac{1}{2}x\right)}\\ &=\frac{2}{1-\cos^2(x)}-\frac{1}{1-\cos(x)}\\ &=\frac{2}{1-\cos^2(x)}-\frac{1+\cos(x)}{1-\cos^2(x)}\\ &=\frac{1-\cos(x)}{1-\cos^2(x)}\\ &=\frac{1}{1+\cos(x)} \end{align*} \]
anonymous
  • anonymous
Another approach using the same identity: \[\begin{align*} 2\csc^2x-\frac{1}{2}\csc^2\frac{x}{2}&=\frac{4}{1-\cos2x}-\frac{1}{1-\cos x}\\[1ex] &=\frac{3-4\cos x+\cos2x}{(1-\cos2x)(1-\cos x)}\\[1ex] &=\frac{2(1-\cos x)^2}{(2-2\cos^2x)(1-\cos x)} \end{align*}\] which gives the last line as in thomas's answer.
ganeshie8
  • ganeshie8
Ahh that identity looks much better as it avoids the half angles/fractions
welshfella
  • welshfella
i didn't know that identity on the second line @ganeshie
welshfella
  • welshfella
that half angle one
welshfella
  • welshfella
I searched the web to find a half angle identity for sin^ x but couldnt find one
ganeshie8
  • ganeshie8
ohh actually it is just the familiar sin(2x) formula, i skipped several steps in between.. let me add them
thomas5267
  • thomas5267
\[ \sin^2\theta = \frac{1 - \cos 2\theta}{2} \] Copied shamelessly from Wikipedia.
welshfella
  • welshfella
Ah i see - you are squaring both sides Right!
ganeshie8
  • ganeshie8
\[\begin{align} 2\csc^2 x - \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(\sin x)^2}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(2\sin\frac{x}{2}\cos\frac{x}{2})^2}-\frac{1}{\sin^2\frac{x}{2}}\right)~~\color{grey}{\because \sin(2x)=2\sin x \cos x}\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1-\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align}\] we can take the limit now
welshfella
  • welshfella
yes I get it now
welshfella
  • welshfella
thanks
ganeshie8
  • ganeshie8
np:) sometimes the answer from wolfram gives some idea on how to approach these limit problems http://www.wolframalpha.com/input/?i=lim%28x%5Cto+0%29+2csc%5E2x-1%2F2csc%5E2%28x%2F2%29

Looking for something else?

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