so didnt pay attention is first calc class a week ago and left the hmwk until tonight.. who can help me "evaluate" the following...
lol good luck
\[\lim_{x \rightarrow 2}(x ^{2}+4x+1)\]
13
so then \[\lim_{h \rightarrow 0}((h-5)^{2}-25) \div h\] is 0
unfortunately no, since the result of that is undefined (0/0) you have to do other stuff to compute the limit (remember a limit doesn't care if the function is defined or not)
so is that when i have to find the "hole"?
do you know l'hopitals rule?
thats probably what went on in your lecture this week!
if hes in the first part of calculus they probably want him to do that w/ algebra and not lhopitals rule (that's what they made me do)
\[\lim_{h \rightarrow 0}\frac{h^2-10h+25-25}{h}=\lim_{h \rightarrow 0}\frac{h^2-10h}{h}=\lim_{h \rightarrow 0}(h-10)\]
yeah that
yes this is newtons approach to taking derivatives! Very good lesson! Calculus I good lord that was long ago!
haha high school calculus, payed attention next calss. that concept was unbelievably easy i must have looked stupid asking that hahahah. wow.
There are no stupid questions.
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