L'hospital rule confusion

**asiler** · December 16th 2007, 07:28 PM

I have the following limit:

$lim$ $t->0$ $(e^t - 1) / t ^ 3$

I use L'hospital's rule to keep differentiating the numerator and denominator, and I end up with 1/6. But, according to my textbook, the correct answer is infinity. Does anyone see what I am doing wrong here?

**ThePerfectHacker** · December 16th 2007, 07:43 PM

$\lim_{t\to 0}\frac{e^t - 1}{t}$ is the derivative of $e^t$ at $t=0$ by definition. Which is $e^0 = 1$ .

That means,
$\lim_{t\to 0}\frac{e^t-1}{t^3} = \lim_{t\to 0}\frac{e^t - 1}{t}\cdot \frac{1}{t^2} = \infty$

**mr fantastic** · December 16th 2007, 10:36 PM

Originally Posted by asiler

I have the following limit:

$lim$ $t->0$ $(e^t - 1) / t ^ 3$

I use L'hospital's rule to keep differentiating the numerator and denominator, and I end up with 1/6. But, according to my textbook, the correct answer is infinity. Does anyone see what I am doing wrong here?

OK, first review under what circumstances l'H can be used. Done it? Right then ......

After the first differentiation you no longer have an indeterminant form of the type 0/0. You're mistake was ........

in the continued use of l'H beyond the first differentiation.

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