Question

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

Answer 1

\[\lim_{x \rightarrow 0^+}\frac{ \sin2x }{ -cosx+sinx }\]

Answer 2

I have already simplified it, the solution says 0. So, how is it 0?

Answer 3

Let's try plugging in x = 0 and see what happens


\[\Large \lim_{x \rightarrow 0^+}\frac{ \sin(2x) }{ -\cos(x)+\sin(x) } =\frac{ \sin(2*0) }{ -\cos(0)+\sin(0) } \]

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

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

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

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

Answer 4

hopefully all that makes sense

Answer 5

Oh, I figured I had to plug in pi/2 or something. Thanks! :)

Answer 6

np