SOLUTION: I'd guess this is an easy problem but something I don't see here. Evaluate the limit lim(t->3) (2^t-8)/(t-3) (Hint: think derivative) Please explain how to solve Thank yo

Algebra ->  Equations -> SOLUTION: I'd guess this is an easy problem but something I don't see here. Evaluate the limit lim(t->3) (2^t-8)/(t-3) (Hint: think derivative) Please explain how to solve Thank yo      Log On


   



Question 995368: I'd guess this is an easy problem but something I don't see here.
Evaluate the limit lim(t->3) (2^t-8)/(t-3) (Hint: think derivative)
Please explain how to solve
Thank you!

Found 3 solutions by addingup, rothauserc, Edwin McCravy:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
Find the limit:
lim_(x->3)(2^t-8)/(t-3)
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Since (2^t-8)/(t-3) is constant with respect to x, and the limit of a constant is that constant, lim_(x->3)(2^t-8)/(t-3)= (2^t-8)/(t-3)= (2^t-8)/(t-3)
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P.S.: you are asking us to find the limit so I'm not sure what you mean by "think derivative". If you try to derivate it you get 0 for an answer:
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d/dx(lim_(x->3)(2^t-8)/(t-3))
Rewrite: lim_(x->3)(2^t-8)/(t-3)= (-8+2^t)/(-3+t):
= d/dx((-8+2^t)/(-3+t))
The derivative of (2^t-8)/(t-3) is zero, therefore your answer is 0

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
did you mean ((2^t)-8)/(t-3) or (2^(t-8))/(t-3)?

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
They don't know what you want.  Here's the solution.



Don't just think derivative, think L'Hopital's rule.  Since both
numerator and denominator approach 0, L'Hopital's rule applies:

To find the derivative of 2t, we use this formula: d%28a%5Eu%29%2Fdx=a%5Eu%2Aln%28u%29%2A%22u%27%22



matrix%282%2C2%2C%0D%0A%0D%0Alim%2C2%5Et%2Aln%282%29%2C%0D%0A%22x-%3E3%22%2C%22%22%29

2%5E3%2Aln%282%29

8ln%282%29

Edwin