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

1 Answer
Steve M
Sep 3, 2017

# 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#

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