How to find the limit of a complicated function using L'Hopital's rule?

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To find the limit of the function as t approaches zero, the term (sin(t)/(2t)) approaches 1/2 using the standard limit of sin(t)/t as t approaches zero. The second term (e^(2t)) approaches j as t approaches zero. The third term (t^2/e^t) approaches zero, contributing nothing to the limit. Therefore, the overall limit is 1/2i + j, not just j. L'Hôpital's rule can be applied to confirm the limit for the first term.
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can someone help me out with a simple calc question?

we are asked to find the limit of a function as t approaches zero. the function is:

(sint/2t)i + (e^2t)j + (t^2/e^t)k

why is the answer not just j?

the answer given is 1/2i + j, but I have no idea how that first term came about.

help!
 
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sin(t)/(2t) as t->0 is one of those indeterminate 0/0 forms, it's not 0 as you seemed to hope. You should know the limit of sin(t)/t as t->0, it's a pretty standard one (or you can use l'hopital's rule).
 
don23 said:
can someone help me out with a simple calc question?

we are asked to find the limit of a function as t approaches zero. the function is:

(sint/2t)i + (e^2t)j + (t^2/e^t)k

why is the answer not just j?

the answer given is 1/2i + j, but I have no idea how that first term came about.

help!
\lim_{t{\to}0}\frac{\sin{t}}{t}=1

edit: i can't remember the diravation at the moment
 
Last edited:
thank you

thank you:) it's been a while...i forgot about l'hopital's rule. thanks again
 
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