Question

Question #4ed65

Answer 1

`= - 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`