Question

lim x tends to 0 sin3X×sin5X ÷ 7X²​

Answer 1

Answer:

In the problem, as x→ 0, the angles 5x→ 0 and 7x→ 0. Therefore, we adjust 5x and 7x in the denominator.

∴ required limit of sin 5x.sin 7x/3x²

= (1/3). lim (sin 5x/x) . lim(sin 7x/x)

= (1/3) . lim (sin 5x/5x) .5. lim (sin 7x/7x) . 7

= (1/3). 1. 5. 1. 7 (Since as x→ 0, 5x→ 0. 7x→ 0 and lim (sin t/t) = 1 as t→ 0)

= (1/3). 5. 7

= 35/3, 11 2/3 or 11.6667 (Answer)