i have to find the limit of this
lim x->0+ $\displaystyle sin(2x)/x^2$
once i simplify this i get $\displaystyle 2/x$
and i get stuck. can someone help? thanks
Yes. $\displaystyle \lim_{x\to0}\frac{\sin{x}}{x}=1$ is a special limit derived for the purposes of determining the trigonometric derivatives.
Check out the proof here...
World Web Math: Useful Trig Limits
kev91: $\displaystyle cos2x \rightarrow 1$ as $\displaystyle x \to 0^+$, not 2, but the answer remains the same.
Another way to do this without L'hopital is $\displaystyle sin(2x) = 2sinxcosx \Rightarrow \frac{sin(2x)}{x^2} = 2 \frac{sinx}{x} \cdot \frac{cosx}{x}$, and now since $\displaystyle \frac{cosx}{x} \to \infty \text{ as } x \to 0^+$, we get $\displaystyle \lim_{x \to 0^+} 2 \cdot \frac{sinx}{x} \cdot \frac{cosx}{x} = 2 \cdot \lim_{x \to 0^+} \frac{cosx}{x} = \infty$
Informally speaking, yes -- when you have a fraction with numerator reaching zero and denominator reaching infinity, the fraction reaches zero. The other statement is true as well : really big number divided by really small number is like really big number times another really big number, equals really really big number!