Question

Continuity of Composite Functions:Evaluate the following limit:
lim sin^2(x)+6sin(x)+5/sin^2x-1 as x approaches 5pi/2

Answer 1

\[\lim_{x \rightarrow 5\pi/2}\frac{ \sin^2x+6sinx+5 }{ \sin^2x-1 }\]
function rewritten.

Answer 2

So far, I know that at x=1 the function is undefined so, now I need to simplify the function and try to find a removeable discontinuity that does not get me zero. This is where I'm having difficulties.

Answer 3

note: 5pi/2 is equivalent to just pi/2 to simplify things

it seems by plugging in pi/2 you get --->
\[\frac{1+6+5}{1-1} = \frac{12}{0} \rightarrow \infty\]

Answer 4

This is what I have so far:

\[\lim_{x \rightarrow 5\pi/2}\frac{ \sin^2x+6sinx+5 }{ \sin^2x-1 }= \frac{ \frac{1-\cos2x }{2}+6sinx+5}{\frac{1-\cos2x }{2}-1}\]

I got 1-cos2x/2 using the hald angle formula. Not sure if I'm on the right track or not. I'm trying to get the answer that wolfram alpha got
http://www.wolframalpha.com/input/?i=factor+%28sin^2x%2B6sinx%2B5%29%2F%28sinx^2x-1%29

Answer 5

http://www.wolframalpha.com/input/?i=lim+%28sin%5E2x%2B6sinx%2B5%29%2F%28sinx%5E2-1%29+as+x-%3Epi%2F2

Answer 6

factorize the numerator and denominator:
\[\lim_{x->5\pi/2} \frac{ \sin^2x + 6sinx + 5 }{ (\sin^2x -1) } = \lim_{x->5\pi/2}\frac{ (sinx + 5)(sinx + 1) }{ (sinx+1)(sinx-1) }\]

Answer 7

Oh I see, so the answer would be sinx+5/sinx+1=3 right?

Answer 8

So if I had a function like this one\[\lim_{x \rightarrow 0^+}cot(x)\], to evaluate this one so that cot(x) doesn't become undefined as x approaches 0^+, can I do this:
\[\lim_{x \rightarrow 0+}\cot(x)=\tan(x)=\sin(x)/\cos(x)=0\]
So 0 would be my answer for this one, is that right as well?