Question

evaluate the limx->0 sin2x/sin5x

with consideration that limx->sinx/x=1

Answer 1

well for starters, it might help to note that:
\[\frac{\sin(2x)}{\sin(5x)} = \frac{\frac{\sin(2x)}{x}}{\frac{\sin(5x)}{x}}\]
i dont know if this is the right way to go, just a suggestion.

Answer 2

\[\lim_{x \rightarrow 0}\frac{\sin2x}{2x}*\frac{5x}{\sin5x}*\frac{2x}{5x}\]

Answer 3

\[\lim_{x \rightarrow 0}1*1*\frac{2x}{5x}\]\[\frac{2x}{5x}\]

Answer 4

i know how to do it but my question is that my lecturer says the answer is 2/5 but substitutiting the x in you get zero? so isnt the answer zero or is it 2/5

Answer 5

but why mathfrick?

Answer 6

you dont substitute anything in for x, just divide them out:
\[\frac{2x}{5x} = \frac{2}{5}(1) = \frac{2}{5}\]

Answer 7

because the x cancel out okay thank you!...i just saw that lol

THANKS SO MUCH JOE!