Question

What is the limit has `h -> 0` of `(3(1/2 +h) ^5 - 3(1/2)^5)/h`?

Answer 1

The limit is `15/16` Here are two approaches. Your preference may depend on your current knowledge of derivatives.

Explanation:

`lim_(hrarr0)(3(1/2+h)^5-3(1/2)^5)/h`

If we try to evaluate by substitution, we see that the numerator and denominator both have limits of `0`. So we cannot use the quotient rule for limits. (We say that the limit has an indeterminate form.)

If you have not yet learned the rules for differentiation , you will evaluate this limit by rewriting the expression.

Note that it may not be clear initially that this approach will work, but we have to try something and this is something we can do.

Expand `(1/2+h)^5`. (Use the binomial theorem -- aka the binomial expansion.)

`(1/2+h)^5 = (1/2)^5+5(1/2)^4h+10(1/2)^3h^2+10(1/2)^2h^3+5(1/2)h^4+h^5`

We will use this to simplify the numerator of the expression of interest.

`(3(1/2+h)^5-3(1/2)^5)/h = (3[(1/2)^5+5(1/2)^4h+10(1/2)^3h^2+10(1/2)^2h^3+5(1/2)h^4+h^5]-3(1/2)^5)/h`

`= (color(red)(3(1/2)^5)+15(1/2)^4h+30(1/2)^3h^2+30(1/2)^2h^3+15(1/2)h^4+3h^5color(red)(-3(1/2)^5))/h`

`= (15(1/2)^4h+30(1/2)^3h^2+30(1/2)^2h^3+15(1/2)h^4+3h^5)/h`

Observe that we can remove a common factor of `h` in the numerator, so we get

`(3(1/2+h)^5-3(1/2)^5)/h = 15(1/2)^4+30(1/2)^3h+30(1/2)^2h^2+15(1/2)h^3+3h^4` for all `h != 0`.

Recall that if `f(x) = g(x)` for all `x` near `a` except possibly at `x=a` then the limits as `xrarra` of the functions either exist together and are equal or neither limit exists.

`lim_(hrarr0)(15(1/2)^4+30(1/2)^3h+30(1/2)^2h^2+15(1/2)h^3+3h^4=15(1/2)^4)=15(1/2)^4`.
Therefore, `lim_(hrarr0)(3(1/2+h)^5-3(1/2)^5)/h = 15(1/2)^4 = 15/16`.

If you have learned the rules for differentiation , there is a faster method.

Notice that the expression `lim_(hrarr0)(3(1/2+h)^5-3(1/2)^5)/h` is exactly the derivative of `f(x) = 3x^5` at `x=1/2`.

So we want `f'(1/2)`.

`f'(x) = 15x^4`, and `f'(1/2) = 15(1/2)^4 = 15/16`.