Question

evaluate `lim_(x rarr 0) (sin3x-sinx)/sinx`?

Answer 1

`lim_(x rarr 0) (sin3x-sinx)/sinx = 2`

Explanation:

We want to evaluate:

`L = lim_(x rarr 0) (sin3x-sinx)/sinx`

We can manipulate the limit as follows:

`L = lim_(x rarr 0) {(sin3x)/sinx-sinx/sinx}`

`\ \ \ = lim_(x rarr 0) sin(x+2x)/sinx-lim_(x rarr 0) sinx/sinx`

`\ \ \ = lim_(x rarr 0) (sinxcos2x+cosxsin2x)/sinx-lim_(x rarr 0) 1`

`\ \ \ = lim_(x rarr 0) {(sinxcos2x)/sinx+(cosxsin2x)/sinx} - 1`

`\ \ \ = lim_(x rarr 0) cos2x +lim_(x rarr 0) (cosxsin2x)/sinx - 1`

`\ \ \ = cos0 + lim_(x rarr 0) (2cosxsinxcosx)/sinx - 1`

`\ \ \ = 1+lim_(x rarr 0) 2cos^2x -1`

`\ \ \ = 2cos^2 0`

`\ \ \ = 2`