Question

limit of x approaches to 0 sin2x÷√5x find a limit​

Answer 1

Answer:

$\frac{2}{\sqrt{5} }$

Explanation:

 $\lim_{x \to 0$   $\frac{sin 2x}{\sqrt{5}x }$

=  $\lim_{x \to 0$ $\frac{sin 2x}{\sqrt{5}x} *\frac{2}{2}$

=  $\lim_{x \to 0$ $\frac{2 sin 2x}{\sqrt{5}*2x}$

=  $\lim_{x \to 0$  $\frac{2}{\sqrt{5} }$

=  $\frac{2}{\sqrt{5} }$