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
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) (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) (Show Source): You can put this solution on YOUR website! did you mean ((2^t)-8)/(t-3) or (2^(t-8))/(t-3)?
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:
Edwin