# L'hospitals rule

1. Nov 20, 2009

### hover

1. The problem statement, all variables and given/known data
evaluate the limit
$$(lim.t\rightarrow0) \frac{e^6^t-1}{t}$$

2. Relevant equations
l'hospital's rule i guess

3. The attempt at a solution
With the usual approach of this rule, you are suppose to take a derivative and evaluate the derivative at that original limit. The problem is that every time I take a derivative and apply the limit I keep getting an indeterminate form. I don't know how to solve this. Can someone point me in the right direction?

Thanks

2. Nov 20, 2009

### Bohrok

Once you use l'Hôpital's rule once, it's not in an indeterminate form and you can find the limit. Try showing your work.

3. Nov 20, 2009

### hover

ok so

$$\frac {d}{dx} (\frac {e^6^t-1}{t}) = \frac{t(6e^6^t)-(e^6^t-1)}{t^2} = \frac {6te^6^t-e^6^t+1}{t^2}$$

apply the limit

$$\frac{0-1+1}{0}= \frac {0}{0}$$

4. Nov 20, 2009

### Bohrok

When using l'Hôpital's rule, you take the derivative of the top and bottom separately, do not use the difference [STRIKE]quotient[/STRIKE] rule!
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)} = \lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}$$

Last edited: Nov 20, 2009
5. Nov 20, 2009

### hover

oh so NO quotient rule? oh ok. let me redo it then hold on.

6. Nov 20, 2009

### hover

well just looking at it the answer is 6. wow thanks for that. I thought you had to use the quotient rule.

many thanks