lim _{x rightarrow 2} frac{x+3 x^2+5 x^3+7 x^ | vedclass

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(D) Let $f(x) = x + 3x^2 + 5x^3 + 7x^4 - 166$. First,check the value of $f(2)$: $f(2) = 2 + 3(2^2) + 5(2^3) + 7(2^4) - 166 = 2 + 12 + 40 + 112 - 166 =

Vedclass · Accepted Answer

$297$

Vedclass · Answer

(D) Let $f(x) = x + 3x^2 + 5x^3 + 7x^4 - 166$. First,check the value of $f(2)$: $f(2) = 2 + 3(2^2) + 5(2^3) + 7(2^4) - 166 = 2 + 12 + 40 + 112 - 166 = 166 - 166 = 0$. Since the limit is of the form $\frac{0}{0}$,we apply $L$'Hopital's Rule: $\lim _{x \rightarrow 2} \frac{f'(x)}{1} = f'(2)$. $f'(x) = \frac{d}{dx}(x + 3x^2 + 5x^3 + 7x^4 - 166) = 1 + 6x + 15x^2 + 28x^3$. Now,evaluate $f'(2)$: $f'(2) = 1 + 6(2) + 15(2^2) + 28(2^3) = 1 + 12 + 15(4) + 28(8) = 1 + 12 + 60 + 224 = 297$. Thus,the limit is $297$.

$\lim _{x \rightarrow 2} \frac{x+3 x^2+5 x^3+7 x^4-166}{x-2} = $

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