MHB Limit Definitions and Finding the Unknown Quick Question

[ardentmed]

Hey guys,

Here's another quick question this time from a problem set I'm having trouble with at the moment.

Question:

So, for a, I computed f(x) = x^5
This is because of the numberator's right side. If 2+h is raised to the 5, this must be the function.

Moreover, a=2 because 2 is already on the right side of the numerator. 2 is added to "x" which confirms 2^5 = 32, also seen in the numerator.

Also, for 1b, I used the definition of the derivative, where
$\lim_{{h}\to{0}}$ (x+h)^5 - x^5 ) / h.

This ultimately gave me 4x^4 as the answer.

Am I on the right track?

Thanks in advance.

 

[mathmari]

Hey guys,

Here's another quick question this time from a problem set I'm having trouble with at the moment.

Question:So, for a, I computed f(x) = x^5
This is because of the numberator's right side. If 2+h is raised to the 5, this must be the function.

Moreover, a=2 because 2 is already on the right side of the numerator. 2 is added to "x" which confirms 2^5 = 32, also seen in the numerator.

Also, for 1b, I used the definition of the derivative, where
$\lim_{{h}\to{0}}$ (x+h)^5 - x^5 ) / h.

This ultimately gave me 4x^4 as the answer.

Am I on the right track?

Thanks in advance.

The definition of the derivative at the point $c$ is
$$\lim_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}=f'(c)$$

So comparing this formula with $$\lim_{h \rightarrow 0} \frac{(2+h)^5-32}{h}=f'(c) \Rightarrow \lim_{h \rightarrow 0} \frac{(2+h)^5-2^5}{h}=f'(c)$$ we see that $f(x)=x^5$ and $c=2$.

To find the limit by evaluating only the derivative we do the following:
$f'(x)=(x^5)'=5x^4$
$$\lim_{h \rightarrow 0} \frac{(2+h)^5-2^5}{h}=f'(2)=5 \cdot 2^4=5 \cdot 16=80$$