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

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best_answer

f\left ( 1 \right )= 9
\lim_{x\rightarrow 1}\frac{x^{n}f\left ( 1 \right )-f\left ( x \right )}{x-1}= 44
= \lim_{x\rightarrow 1}\frac{9x^{n}-f\left ( x \right )}{x-1}= 44

For this limit to exist, numerator must tend to zero

For \frac{0}{0} Limit, Apply L'Hospital Rule
\Rightarrow \lim_{x\rightarrow 1}\frac{9nx^{n-1}-{f}'\left ( x \right )}{1}= 44
\Rightarrow 9n-{f}'\left ( 1 \right )= 44

As\,\,f \,'\left ( x \right )= 6x^{5}+8x^{3}+3x^{2}+2
{f}'\left ( 1 \right )= 19

\Rightarrow 9n-19= 44\Rightarrow 9n= 63\Rightarrow n= 7

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Kuldeep Maurya

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