Question #b5c34

2 Answers
Sep 16, 2017

Answer:

We seek:

# lim_(x rarr 0) (tan2x-sin2x)/x^2 = 0#

Explanation:

We seek:

# L = lim_(x rarr 0) (tan2x-sin2x)/x^2#

If we put #x=09# we note both numerator and denominator are both #0#, so we have an indeterminate form #0/0#, and as such we can apply L'Hôpital's Rule to give:

# L = lim_(x rarr 0) (d/dx(tan2x-sin2x))/(d/dx x^2) #
# \ \ \ = lim_(x rarr 0) (2sec^2x-2cos2x)/(2x) #
# \ \ \ = lim_(x rarr 0) (sec^2x-cos2x)/(x) #

Again, we have an indeterminate form #0/0#, and as such we can apply L'Hôpital's Rule again to give:

# L = lim_(x rarr 0) (d/dx(sec^2x-cos2x))/(d/dx x) #
# \ \ \ = lim_(x rarr 0) (2secxsecxtanx+2sinx)/(1) #
# \ \ \ = 2lim_(x rarr 0) 2secx^2tanx+sinx #
# \ \ \ = 2(0+0)#
# \ \ \ = 0#

Sep 16, 2017

Answer:

# 0.#

Explanation:

Let us find the Limit L without using L'Hospital's Rule.

We will use #lim_(theta to 0) tantheta/theta=1.#

Knowing that, #tan 2x=(2tanx)/(1-tan^2x), &, sin2x=(2tanx)/(1+tan^2x),#

we have,

#tan2x-sin2x,#

#=(2tanx)/(1-tan^2x)-(2tanx)/(1+tan^2x),#

#=2tanx{{(1+tan^2x)-(1-tan^2x)}/((1-tan^2x)(1+tan^2x)}},#

#=(4tan^3x)/(1-tan^4x)xxx^3/x^3,#

#=4*(tanx/x)^3*(x^2*x)/(1-tan^4x).#

# rArr (tan2x-sin2x)/x^2=4*(tanx/x)^3*x/(1-tan^4x).#

#:. L=lim_(x to 0)4*(tanx/x)^3*x/(1-tan^4x),#

#=4*1^3*0/(1-0^4),#

# rArr L=0.#

Enjoy Maths.!