Evaluate the limit
(6+h)^3-216/h
I think you would have expand the numerator
(6+h)^3 = but how would you do it?
216+108h+h^3? It's not coming out right...
(6+h)^3
= 216 + 3(36h) + 3(6h^2) + h^3
= 216 + 108h + 18h^2 + h^3
so Limit ( (6+h)^3-216 )/h , as h ---> 0
= lim ( 216 + 108h + 18h^2 + h^3 - 216)/h
= lim (108h + 18h^2 + h^3)/h , as h --> 0
= lim 108 + 18h + h^2 , as h --> 0
= 108
check using Calculus as such ...
we were finding dy/dx of
y = x^3, when x = 6
dy/dx = 3x^2
= 3(6^2) = 108
To expand the numerator term (6+h)^3 correctly, you can use the binomial expansion formula or the concept of the distributive property of exponents.
The binomial expansion formula states that (a+b)^n can be expanded using the binomial coefficients and the powers of a and b. In this case, a is 6 and b is h, and n is 3.
The formula is:
(a+b)^n = C(n,0)*a^n*b^0 + C(n,1)*a^(n-1)*b^1 + C(n,2)*a^(n-2)*b^2 + ... + C(n,n-1)*a^1*b^(n-1) + C(n,n)*a^0*b^n
Therefore, we can substitute the values a=6, b=h, and n=3 into the formula:
(6+h)^3 = C(3,0)*6^3*h^0 + C(3,1)*6^2*h^1 + C(3,2)*6^1*h^2 + C(3,3)*6^0*h^3
Now, let's calculate the binomial coefficients C(n,k). The binomial coefficient C(n,k) is equal to n!/(k!(n-k)!), where "!" denotes the factorial function.
C(3,0) = 3!/(0!(3-0)!) = 1
C(3,1) = 3!/(1!(3-1)!) = 3
C(3,2) = 3!/(2!(3-2)!) = 3
C(3,3) = 3!/(3!(3-3)!) = 1
Now substituting these values back into (6+h)^3:
(6+h)^3 = 1*6^3*h^0 + 3*6^2*h^1 + 3*6^1*h^2 + 1*6^0*h^3
Simplifying further:
(6+h)^3 = 216 + 108h + 18h^2 + h^3
So, the correct expansion of (6+h)^3 is 216 + 108h + 18h^2 + h^3.