Question #4ed65

Feb 13, 2017

$= - \frac{3}{2}$

Explanation:

For ${\lim}_{x \to 0} \frac{1 - {e}^{3 x}}{\sin \left(2 x\right)}$, if we plug $x = 0$ straight in, we see that this is in indeterminate form: $= \frac{1 - {e}^{0}}{\sin \left(0\right)} = \frac{1 - 1}{0} = \frac{0}{0}$.

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

$= {\lim}_{x \to 0} \frac{- 3 {e}^{3 x}}{2 \cos \left(2 x\right)}$

If we plug $x = 0$ in, this is now: $\frac{- 3 {e}^{0}}{2 \cos \left(0\right)} = - \frac{3}{2}$.

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

• ${e}^{z} = 1 + z + \setminus O \left({z}^{2}\right)$

• $\sin z = z + \setminus O \left({z}^{2}\right)$

Sub'ing these into: ${\lim}_{x \to 0} \frac{1 - {e}^{3 x}}{\sin \left(2 x\right)}$, gives us this:

${\lim}_{x \to 0} \frac{1 - \left(1 + 3 x + \setminus O \left({x}^{2}\right)\right)}{\left(2 x\right) + \setminus O \left({x}^{2}\right)}$

$= {\lim}_{x \to 0} \frac{- 3 x + \setminus O \left({x}^{2}\right)}{2 x + \setminus O \left({x}^{2}\right)}$

$= {\lim}_{x \to 0} \frac{- 3 + \setminus O \left(x\right)}{2 + \setminus O \left(x\right)}$

$= - \frac{3}{2}$