Question

plzz choose correct option

Answer 1

$Given : \lim_{n \to 0} \frac{sin 5x}{x}$

By L-hospital rule, we get

$\lim_{n \to 0} \frac{(cos 5x) * 5}{1}$

$\lim_{n \to 0} (5 cos 5x)$

5 cos(5 * 0)

5 cos 0

5 * 1

5.


Hope this helps!

Answer 2

heya...!!!

we know that:-

$lim \frac{sinx}{x} = 1 \\ x - > 0$

now,


$lim \frac{sin5x}{x} \\ x - > 0 \\ \\ = > lim \frac{(5 \times sin5x)}{(5 \times x)} \\ x > 0 \\ \\ = > lim (\frac{5 \times sin5x}{5x} ) \\ x - > 0 \\ \\ = > 5 \times 1 = 5$

hence option (c) is correct