Question

I have a calculus question about limits and derivatives. pic included

Answer 1

Step-by-step explanation

The limit definition of derivative is:

`f^(\prime)(x)=\lim_(h\to0)(f(x+h)-f(x))/(h)`

The given function is a constant function, which means that no matter what value is assigned to h, f(x+h) will always be equal to 3.

`\begin{gathered} f^(\prime)(x)=\lim_(h\to0)(f(x+h)-f(x))/(h) \\ f^(\prime)(x)=\lim_(h\to0)(3-3)/(h) \\ f^(\prime)(x)=\lim_(h\to0)(0)/(h) \end{gathered}`

Now, we evaluate the limit of the numerator and the denominator.

`\begin{gathered} f^(\prime)(x)=(\lim_(h\to0)0)/(\lim_(h\to0)h) \\ f^(\prime)(x)=(0)/(0) \end{gathered}`

Since we have an indeterminate form, we apply the L'Hospital's Rule which states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

`\begin{gathered} f^(\prime)(x)=\lim_(h\to0)(0)/(h) \\ f^(\prime)(x)=\lim_(h\to0)((d)/(dh)0)/((d)/(dh)h) \\ f^(\prime)(x)=\lim_(h\to0)(0)/(1) \\ f^(\prime)(x)=\lim_(h\to0)0 \end{gathered}`

Finally, the limit as h approaches 0 is 0.

`\begin{gathered} f^(\prime)(x)=\lim_(h\to0)0 \\ f^(\prime)(x)=0 \end{gathered}`Answer
`f^(\prime)(x)=0`