Medical Malpractice

L- Hospitals Rule CALC 1?

This is one more i am stuck on. L- Hospitals Rule Find the limlt. Use L'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If L'Hospital's Rule doesn't apply, expla!n why. 1. lim t-->0 for ( e^(3t) - 1 ) / t

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1. Lim = d/dx [e^(3t)-1] / d/dx (t) t --> 0 use the chain rule: let u = 3t then d/du e^u = e^u d/dx (3t) = 3 lim = 3e^(3t) t --> 0 lim = 3e^(3*0) t --> 0 lim = 3 <== answer t --> 0
2. First, plug in 0 for t and see what you get. ( e^(3*0) - 1) / 0 = (1-1) / 0 = 0/0 We have 0/0 indeterminate form, so use L'Hopital's Rule and differentiate the top and the bottom. d/dx (e^(3t)-1) / d/dx (t) = 3e^(3t) / 1 = 3e^(3t) Now plug in 0 again. 3e^(3*0) = 3
3. Ok lim e^(3t) - 1 as t ----> 0 = e^(0) - 1 = 1 - 1 = 0 lim (e^(3t) - 1) / t as t -----> 0 = 0 / 0 which is undefined, L'hospital rule is applicable here (Remember that this rule is only applicable when you have a 0 /0 format or infinity / infinity). The limit becomes: lim 3e^(3t) / 1 as t --->0 = 3e^0 = 3 So lim (e^(3t) - 1) / t as t ----> 0 = 3
4. L'Hospital's rule says that given a function h(x) = f(x)/g(x) where f(x) -> 0 and g(x) -> 0 when x-> 0, you can find the limit by evaluating f'(x)/g'(x) as x->0 where the ' menas derivative with respect to x. SO let's use this rule: lim t->0 (e^(3t)-1)/t as t-> 0 e^(3t)-1 ->0 and t->0 so we can use the rule. d(e^(3t)-1)/dt = 3e^(3t) and d(t)/dt = 1 Thus: lim t->0 (e^(3t)-1)/t = lim t -> 0 3e^(3t)/1 = 3