Evaluate the limit #lim_(x rarr 0) sin(x)/sin(4x)#?
3 Answers
Explanation:
We will use the Standard Limit
Now, the Reqd. Lim.
Explanation:
When solving this limit, the first step is to use direct substitution which looks like this
This answer is indeterminate and therefore we would use L'Hôpital's rule which says to treat the numerator and denominator as separate functions and to take the derivative of each one like this
L'Hôpital's Rule:
This case:
Use the same limit
Substitute the value of the limit into the new expression
There is your limit
# lim_(x rarr 0) sinx/sin(4x) = 1/4 #
Explanation:
Using the trig identity:
# sin 2A -= 2sinAcosA #
We have:
# lim_(x rarr 0) sinx/sin(4x) = lim_(x rarr 0) sinx/sin(2(2x))#
# " " = lim_(x rarr 0) sinx/(2sin2xcos2x)#
# " " = lim_(x rarr 0) sinx/(2(2sinxcosx)cos2x)#
# " " = lim_(x rarr 0) sinx/(4sinxcosxcos2x)#
# " " = lim_(x rarr 0) 1/(4cosxcos2x)#
# " " = 1/(4*1*1)#
# " " = 1/4 #