Limits: What is the limit of ((2+h)^4 - 16)/h as h approaches 0?

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Homework Statement


Evaluate
lim ((2+h)^4 - 16)/h
h->0



Homework Equations


Difference of squares


The Attempt at a Solution


knowing that the top of the fraction is a difference of squares
i factored it and arrived at lim h->0 ((2+h)^2+4)((2+h)^2-4)/h
this is where i got stuck
 
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Try multiplying the numerator expression to simplify instead of factoring it.
 
Now just expand both (2+h)^{2} as well, and then distribute what's left in the numerator. If you completely simplify the whole thing to expand that giant polynomial at the top, you might be able to cancel that h in the denominator and take the limit.
 
well, i think i followed your method quark which was to expand the (2+h)^2
i ended up (after simplification) with lim h->0 (h^2 + 4h + 8)(h^2 + 4h)/h but i do not know how to expand that. can you please show me how to arrive to the answer?

the answer in the back of the textbook is 32.

also, is this the easiest way to solve the problem?
 
From:
\frac{(h^{2}+4h+8)(h^{2}+4h)}{h}
You can distribute the numerator. It works the same way as if you were going to distribute/foil something like (a+b)(c+d)=(ac+da+bc+dc)

In this case it would be something more like this:
(a+b+c)(d+e+f) = (ad+ae+af+bd+be+bf+cd+ce+cf)

I think the easiest way of solving this problem (without using any calculus other than the limit) would be to expand the ()^4 right from the start. It's important to know how to distribute like shown above though.
 
QuarkCharmer said:
From:
\frac{(h^{2}+4h+8)(h^{2}+4h)}{h}
You can distribute the numerator. It works the same way as if you were going to distribute/foil something like (a+b)(c+d)=(ac+da+bc+dc)

In this case it would be something more like this:
(a+b+c)(d+e+f) = (ad+ae+af+bd+be+bf+cd+ce+cf)

I think the easiest way of solving this problem (without using any calculus other than the limit) would be to expand the ()^4 right from the start. It's important to know how to distribute like shown above though.

alright, I've arrived at my answer :)
thank you very much
 
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