Question

The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.

विकल्प

• 0.00

• 1.00

• 2.00

• 3.00

Answer 1

The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is 1.00.

Explanation:

`lim_(x→0) (sinx)^2/((e^(tan^2x) - 1))`

= `lim_(x→0)(sinx/x)^2 xx (tan^2x)/((e^(tan^2x) - 1)) xx (x/tanx)^2` = 1