Question #4ed65

1 Answer
Feb 13, 2017

Answer:

#= - 3/2#

Explanation:

For #lim_(x to 0) (1-e^(3x))/(sin(2x))#, if we plug #x = 0# straight in, we see that this is in indeterminate form: #= (1-e^(0))/(sin(0)) = (1-1)/0 = 0/0#.

We can therefore apply L'Hôpital's Rule , which on the first pass yields this:

#= lim_(x to 0) (-3e^(3x))/(2cos(2x))#

If we plug #x = 0# in, this is now: # (-3e^(0))/(2cos(0)) = -(3)/2 #.

We can shed some light on this by looking at the (curtailed) Taylor Expansions for #e^z# and #sin z#:

  • #e^{z}= 1+z+\O(z^2)#

  • #sin z = z + \O(z^2)#

Sub'ing these into: #lim_(x to 0) (1-e^(3x))/(sin(2x))#, gives us this:

#lim_(x to 0) (1-(1 + 3x + \O(x^2)))/((2x) + \O(x^2)) #

#= lim_(x to 0) (- 3x + \O(x^2))/(2x + \O(x^2))#

#= lim_(x to 0) (- 3 + \O(x))/(2 + \O(x))#

#= - 3/2#