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  4. If l= displaystyle lim x arrow 3 (x3-7 x2+15 x-9/x4-5 x3+27 x-27), then find the value of 36 l-7.

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Q. If $l=\displaystyle\lim _{x \rightarrow 3} \frac{x^{3}-7 x^{2}+15 x-9}{x^{4}-5 x^{3}+27 x-27}$, then find the value of $36 l-7$.

Limits and Derivatives

Solution:

$l =\displaystyle\lim _{x \rightarrow 3} \frac{x^{3}-7 x^{2}+15 x-9}{x^{4}-5 x^{3}+27 x-27}$
$=\displaystyle\lim _{x \rightarrow 3} \frac{(x-3)\left(x^{2}-4 x+3\right)}{(x-3)\left(x^{3}-2 x^{2}-6 x+9\right)}$
$=\displaystyle\lim _{x \rightarrow 3} \frac{x^{2}-4 x+3}{x^{3}-2 x^{2}-6 x+9}$
$=\displaystyle\lim _{x \rightarrow 3} \frac{(x-1)(x-3)}{(x-3)\left(x^{2}+x-3\right)}$
$=\displaystyle\lim _{x \rightarrow 3} \frac{x-1}{x^{2}+x-3}=\frac{2}{9}$
$\Rightarrow 36 l-7 =36\left(\frac{2}{9}\right)-7$
$=8-7=1$