Basics of trigonometric limits!

As $x$ approaches 0, we have $\frac{3x+2x}{3x-2x} = \frac{3+2}{3-2} = \boxed{5}$ . Remember that for extremely small $x$ , both $sin x$ and $tan x$ approach $x$ .

this problem is very easy.. divide numerator and denominater by x.then use standard limits sin(nx) by x as x tends to zero..

This is level 5...?

I just applied L-H rule once to get that..

The above limit can be written as:

$\frac{\frac{d}{dx}(sin3x+sin2x)}{\frac{d}{dx}(tan3x-tan2x)}$

$\frac{(3cos3x+2cos2x)}{(3sec²3x-2sec²2x)}$

Which on applying limit x --> 0 gives (3+2)/(3-2) = 5