Question #eefeb

2 Answers
Sep 28, 2015

Answer:

By using trig identities and algebraic manipulation, together with L'Hospitals Rule and other limit laws, we eventually get this limit equal to #1/2#.

Explanation:

By a trig identity, #sin2x=2sinxcosx#

Taking out a common denominator and simplifying algebraically, we get

#lim_(x->0)[x/(2sinxcosx)-sinx] = lim_(x->0)((x-2sin^2xcosx)/(2sinxcosx))#

This is an indeterminant limit form of type #0/0# and therefore we may use L'Hospital's Rule to evaluate it to obtain

#lim_(x->0)((x-2sin^2xcosx)/(2sinxcosx))= lim_(x->0)[(d/dxx-2sin^2xcosx)/(d/dx2sinxcosx)]#

#=lim_(x->0)((1-sin^3x-4sinxcos^2x)/(2cos^2x-2sinxcosx))#

#=1/2#

Sep 28, 2015

Answer:

See the explanation section.

Explanation:

To find #lim_(xrarr0)(x/sin(2x) - sinx)#,

we'll need the fundamental trigonometric limit: #lim_(thetararr0)sintheta/theta = 1#

in the reciprocal form:
#lim_(thetararr0)theta/sintheta = 1#

#lim_(xrarr0)(x/sin(2x) - sinx) = lim_(xrarr0)(x/sin(2x)) - lim_(xrarr0)sinx #

# = lim_(xrarr0)(x/sin(2x))-0#

So the challenge is finding #lim_(xrarr0)x/sin(2x)#

Solution 1
#lim_(xrarr0)x/sin(2x) = lim_(xrarr0)1/2 (2x)/sin(2x)#

# = 1/2 lim_(xrarr0) ((2x))/sin(2x)#

# = 1/2 (1) = 1/2#

We used the fact that as #xrarr0#, we also have #2xrarr0#. We could formalize the substitution #theta = 2x# is we want to be precise.

(The technique in solution 1 also works for #lim_(xrarr0)x/sin(5x) = 1/5#.)

Solution 2

Use #sin(2x) = 2sinxcosx# to rewrite.

#lim_(xrarr0)x/sin(2x) = lim_(xrarr0)x/(2sinxcosx)#

# = lim_(xrarr0)x/(2sinxcosx)#

# = lim_(xrarr0)1/2 x/sinx 1/cosx#

# = 1/2[lim_(xrarr0)x/sinx][lim_(xrarr0)1/cosx]# (if both limits exist)

# = 1/2[1][1/1] = 1/2#

(Unless you know the formula for #sin(5x)#, this solution method will not work for #lim_(xrarr0)x/sin(5x)#.)