Question

Find the numerical value for W such that l'Hospital's Rule applies to the following limit and then compute the value, L, of that limit. limx→1 (2x^2+6x+W)/(x−1)=L

Answer 1

L'Hopital will apply to an indeterminate form of    $\bf \cfrac{\infty}{\infty}\quad or\quad \cfrac{0}{0}$

so.. in this case.. let's say hmmm x-1 at the bottom, turns to 0, since it ends up as 1-1

so.. let's do the same for the numerator, so 2x²+6+w... well.. well, if we set x = 1, we get 2(1)²+6(1)+w.... now, 2(1)² + 6(1) is 8, so, if we just set w = -8, we'd end up with 8 - 8 or 0 atop as well, in which case, L'Hopital would apply

$\bf \lim\limits_{x\to 1}\ \cfrac{2x^2+6x-8}{x-1}\implies \underline{LH}\quad \lim\limits_{x\to 1}\ \cfrac{4x+6}{1}\implies 10$