How do you evaluate the limit #x/sin(x/2)# as x approaches #0#?

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Answer 1 · 2016-09-10T20:34:48

#2#

Our destination is well-known limit usually written as #lim_(z to 0) (sin z)/z = 1# but we can flip it: #lim_(z to 0) z/(sin z) = 1 quad triangle#

So

#lim_(x to 0) x/sin(x/2) #

#= lim_(x to 0) 2 (x/2)/sin(x/2) #

With sub #y = x/2# and lifting the constant outside the limit

#=2 lim_(y to 0) y/sin y #

Now see #triangle#

#=2 * 1 = 2#