Question

Need help with this one:-
Find limit as x --> 0 of 2csc^2 x - (1/2) csc^2 (1/2) x.

Answer 1

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.

Answer 2

i think we can try to simplify the expression using trig identities

Answer 3

\[\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

Answer 4

yea

Answer 5

1/2

Answer 6

Not really awake. Use with caution.
\[
\begin{align*}
&\phantom{=}2\csc^2(x)-\frac{1}{2}\csc^2\left(\frac{1}{2}x\right)\\
&=\frac{2}{\sin^2(x)}-\frac{1}{2\sin^2\left(\frac{1}{2}x\right)}\\
&=\frac{2}{1-\cos^2(x)}-\frac{1}{1-\cos(x)}\\
&=\frac{2}{1-\cos^2(x)}-\frac{1+\cos(x)}{1-\cos^2(x)}\\
&=\frac{1-\cos(x)}{1-\cos^2(x)}\\
&=\frac{1}{1+\cos(x)}
\end{align*}
\]

Answer 7

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.

Answer 8

Ahh that identity looks much better as it avoids the half angles/fractions

Answer 9

i didn't know that identity on the second line @ganeshie

Answer 10

that half angle one

Answer 11

I searched the web to find a half angle identity for sin^ x but couldnt find one

Answer 12

ohh actually it is just the familiar sin(2x) formula, i skipped several steps in between.. let me add them

Answer 13

\[
\sin^2\theta = \frac{1 - \cos 2\theta}{2}
\]
Copied shamelessly from Wikipedia.

Answer 14

Ah i see - you are squaring both sides Right!

Answer 15

\[\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

Answer 16

yes I get it now

Answer 17

thanks

Answer 18

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