Let f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3,x | vedclass

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(B) Given $f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3$. First,calculate $f(1) = 1^{6} + 2(1)^{4} + 1^{3} + 2(1) + 3 = 1 + 2 + 1 + 2 + 3 = 9$. The limit is

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$7$

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(B) Given $f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3$. First,calculate $f(1) = 1^{6} + 2(1)^{4} + 1^{3} + 2(1) + 3 = 1 + 2 + 1 + 2 + 3 = 9$. The limit is $\lim_{x \rightarrow 1} \frac{x^{n} f(1) - f(x)}{x - 1} = 44$. Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ as $x \rightarrow 1$. $1^{n} f(1) - f(1) = 9 - 9 = 0$. Applying $L$'$H$ôpital's Rule: $\lim_{x \rightarrow 1} \frac{n x^{n-1} f(1) - f'(x)}{1} = 44$. Calculate $f'(x) = 6x^{5} + 8x^{3} + 3x^{2} + 2$. At $x = 1$,$f'(1) = 6 + 8 + 3 + 2 = 19$. Substituting these values: $n(1)^{n-1}(9) - 19 = 44$. $9n - 19 = 44$. $9n = 63$. $n = 7$.

Let $f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3$,$x \in R$. Then the natural number $n$ for which $\lim_{x \rightarrow 1} \frac{x^{n} f(1) - f(x)}{x - 1} = 44$ is ...... .

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