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.<br /><br />First, let's find the derivative of \( f(x) \):<br />\[ f'(x) = \frac{d}{dx}(x^8 + x + 8) = 8x^7 + 1 \]<br /><br />Next, we need to evaluate \( f'(-1) \):<br />\[ f'(-1) = 8(-1)^7 + 1 = -8 + 1 = -7 \]<br /><br />Now, we can set up the expression for the limit:<br />\[ \lim_{w \to -1} \frac{f'(w) - f'(-1)}{w + 1} \]<br /><br />Substitute \( f'(w) \) and \( f'(-1) \) into the expression:<br />\[ \lim_{w \to -1} \frac{8w^7 + 1 - (-7)}{w + 1} = \lim_{w \to -1} \frac{8w^7 + 8}{w + 1} \]<br /><br />Factor out the common term in the numerator:<br />\[ \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} \]<br /><br />Notice that \( w^7 + 1 \) can be factored using the sum of cubes:<br />\[ w^7 + 1 = (w^2 - w + 1)(w^5 + w^4 + w^3 + w^2 + w + 1) \]<br /><br />However, for the limit, we can directly substitute \( w = -1 \) into the simplified form:<br />\[ \lim_{w \to -1} \frac{8(w^7 + 1)}{w + 1} = \frac{8((-1)^7 + 1)}{-1 + 1} = \frac{8(0)}{0} \]<br /><br />Since this form is indeterminate, we should simplify further:<br />\[ \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 +