Question #4ed65

1 Answer
Eddie
Feb 13, 2017

= - 3/2=−32

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

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