l'hospitals rule

hover

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


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

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.

hover

ok so

[tex] \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}[/tex]

apply the limit

[tex] \frac{0-1+1}{0}= \frac {0}{0} [/tex]

Bohrok

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

hover

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

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