Evaluate? #lim/(xrarr0)##((2+h)^3-8)/h#

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Answer 1 · 2018-02-03T21:29:48

#lim_(h->0)(((2+h)^3-8)/h)=12#

Hmm... this looks similar to the definition of a derivative:

#lim_(h->0)((f(x+h)-f(x))/h)#

In #((2+h)^3-8)/h#, we seem to be cubing a given value of #x#.
Therefore, we say that #f(x)=x^3#.

We can rewrite our problem:
#lim_(h->0)(((2+h)^3-8)/h)#

=>#lim_(h->0)(((2+h)^3-2^3)/h)#

Now, our problem seems to ask:
"What is the instantaneous rate of change of the function #f(x)=x^3# when #x=2#?"

Instead of trying out this limit, we can now calculate the derivative of the function #f(x)=x^3# when #x=2# using the power rule.

Power rule states that: If #f(x)=x^n# then #f'(x)=nx^(n-1)# where #n# is a constant.

We apply the rule to our situation:
Since #f(x)=x^3#,

#f'(x)=3x^2#

=>#f'(2)=3(2^2)#

=>#f'(2)=12#

That is our answer!