Question #eefeb
2 Answers
By using trig identities and algebraic manipulation, together with L'Hospitals Rule and other limit laws, we eventually get this limit equal to
Explanation:
By a trig identity,
Taking out a common denominator and simplifying algebraically, we get
This is an indeterminant limit form of type
See the explanation section.
Explanation:
To find
we'll need the fundamental trigonometric limit:
in the reciprocal form:
# = lim_(xrarr0)(x/sin(2x))-0#
So the challenge is finding
Solution 1
# = 1/2 lim_(xrarr0) ((2x))/sin(2x)#
# = 1/2 (1) = 1/2#
We used the fact that as
(The technique in solution 1 also works for
Solution 2
Use
# = 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