Difficult Calculus II integral

[QuarkCharmer]

Homework Statement

$$\int_{-1}^{0} \frac{e^{\frac{1}{x}}}{x^{3}}dx$$
Solve the integral and determine if it converges/diverges ect.

Homework Equations

The Attempt at a Solution

$$\int_{-1}^{0} \frac{e^{\frac{1}{x}}}{x^{3}}dx$$
$$\lim_{t \to 0}\int_{-1}^{t} \frac{e^{\frac{1}{x}}}{x^{3}}dx$$

$u = \frac{1}{x}$
$du = \frac{1}{x^{2}}dx$

$$\lim_{t \to 0}\int_{-1}^{\frac{1}{t}} \frac{e^{u}(-x^{2})}{x^{3}}du$$
$$\lim_{t \to 0}\int_{-1}^{\frac{1}{t}} \frac{-e^{u}}{x} du$$
$$x = \frac{1}{u}$$
$$\lim_{t \to 0}\int_{-1}^{\frac{1}{t}} \frac{-e^{u}}{\frac{1}{u}} du$$
$$\lim_{t \to 0} \int_{-1}^{\frac{1}{t}} -e^{u}u du$$

$v = u$
$dv = du$
$w = -e^{u}$
$dw = -e^{u}du$

$$\lim_{t \to 0} (-e^{u}u)_{-1}^{\frac{1}{t}} - \int_{-1}^{\frac{1}{t}} -e^{u}du$$

$$\lim_{t \to 0} (\frac{-e^{\frac{1}{t}}}{t} + e^{-1}(-1)) - (-e^{\frac{1}{t}}+e^{-1})$$

$$\lim_{t \to 0} (\frac{-e^{\frac{1}{t}}}{t}-e^{-1}) + e^{\frac{1}{t}}-e^{-1}$$

$$\lim_{t \to 0} (e^{\frac{1}{t}}-\frac{e^{\frac{1}{t}}}{t}-2e^{-1})$$

And now I can't solve this limit. What am I doing wrong here? I put the limit into my TI-89 and as I thought it is undefined. I know the solution is that it converges to $\frac{-2}{e}$ but I can't seem to get there.

 

[SammyS]

The indefinite integral, $\displaystyle \int\, \frac{e^{\frac{1}{x}}}{x^{3}}dx$

becomes $\displaystyle -e^{u}u - \int\, -e^{u}du\,,$ where u = 1/x

$\displaystyle = -e^{u}u + e^u<br /> +C\,.$

Back substituting (u=1/x) gives:

$\displaystyle \int\, \frac{e^{\frac{1}{x}}}{x^{3}}dx=-\frac{e^{1/x}}{x}+e^{1/x}+C$

Now, evaluate $\displaystyle \lim_{t \to 0}\int_{-1}^{t} \frac{e^{\frac{1}{x}}}{x^{3}}dx \,.$