Problem Set:Problem (1 point) Let f(x)=x^8+x+8 Find lim _(warrow -1)(f'(w)-f'(-1))/(w+1)= square

To solve this problem, we need to find the derivative of the function \( f(x) = x^8 + x + 8 \) and then evaluate the limit.First, let's find the derivative of \( f(x) \): Next, we need to evaluate \( f'(-1) \): Now, we can set up the expression for the limit: Substitute \( f'(w) \) and \( f'(-1) \) into the expression: Factor out the common term in the numerator: Notice that can be factored using the sum of cubes: However, for the limit, we can directly substitute into the simplified form: Since this form is indeterminate, we should simplify further:\[ \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \lim_{w \to -1} \frac{8(w^7 +