Question

lim (x+sinx)/(sin2x) as x approaches 0 .... can anyone help me please ?

Answer 1

There are a lot of ways to evaluate this limit. If you are in your first calculus course, you've just learned that \lim_{t \to 0} \sin(t) /t=1.

Using that, in this case we have that
\lim_{x \to 0} \sin(2x)/x=
2 \lim_{x \to 0} \sin(2x) / (2x) =
2 * 1=2.

Next semester, you'll learn L'Hospital's Rule. Basically, L'Hospital's rule says that if f(x)/g(x) is an indeterminate form of type 0/0 or infinity/infinity, then \lim_{x \to a} f(x)/g(x)=\lim_{x \to a} f'(x)/g'(x).

Applying L'Hospital's rule to this limit gives us

\lim_{x \to 0} \sin(2x)/x =
\lim_{x \to 0} 2 \cos(2x)/1=
2*1/1=2.

Another approach would be to divide the Maclaurin series for sin(2x) by x and then evaluate the limit of the sequence as x approaches 0. (The first term of the series Maclaurin series for sin(2x) /x is 2 which shows that the limit is 2.)

So, there are lots of ways to compute the limit. If you are in Calculus I, your teacher is expecting to see the first approach.