Question

Evaluate the given expression : $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 6x + 8}}$ .

Answer 1

It is given in the question that Evaluate $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 6x + 8}}$ .
Firstly, we will pass the limit to check whether the given question is the form of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ . If this is the case then differentiate both the numerator and denominator and pass the limit to get the required answer.

Complete step-by-step answer:
It is given in the question that Evaluate $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 6x + 8}}$.
Now, pass the limit in the equation, we get,
Therefore, $\dfrac{{{2^3} - 6{{\left( 2 \right)}^2} + 11\left( 2 \right) - 6}}{{{2^2} - 6\left( 2 \right) + 8}}$
$= \dfrac{{8 - 6\left( 4 \right) + 22 - 6}}{{4 - 12 + 8}}$
$= \dfrac{{8 - 24 + 22 - 6}}{{4 - 12 + 8}}$
$= \dfrac{0}{0}$
Since, we got the answer of $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 6x + 8}}$ after passing the limit as $\dfrac{0}{0}$ . So, we will differentiate both the numerator and denominator then pass the limit.
$= \mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\dfrac{d}{{dx}}\left( {{x^3} - 6{x^2} + 11x - 6} \right)}}{{\dfrac{d}{{dx}}\left( {{x^2} - 6x + 8} \right)}}} \right)$
$= \mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{\dfrac{d}{{dx}}{x^3} - \dfrac{d}{{dx}}6{x^2} + \dfrac{d}{{dx}}11x - \dfrac{d}{{dx}}6}}{{\dfrac{d}{{dx}}{x^2} - \dfrac{d}{{dx}}6x + \dfrac{d}{{dx}}8}}} \right)$
$= \mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{3{x^2} - 12x + 11}}{{2x - 6}}} \right)$
Now, passing the limit $x = 2$
$= \left( {\dfrac{{3{{\left( 2 \right)}^2} - 12\left( 2 \right) + 11}}{{2\left( 2 \right) - 6}}} \right)$
$= \left( {\dfrac{{3\left( 4 \right) - 24 + 11}}{{4 - 6}}} \right)$
$= \left( {\dfrac{{12 - 24 + 11}}{{4 - 6}}} \right)$
$= \left( {\dfrac{{ - 1}}{{ - 2}}} \right)$
$= \dfrac{1}{2}$
Hence, $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 6x + 8}} = \dfrac{1}{2}$.

Note: The above question can be solved by using another method i.e. method of factorization.
It is given in the question that Evaluate $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 6x + 8}}$ .
$= \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - {x^2} - 5{x^2} + 5x + 6x - 6}}{{{x^2} - 4x - 2x + 8}}$
$= \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2}\left( {x - 1} \right) - 5x\left( {x - 1} \right) + 6\left( {x - 1} \right)}}{{x\left( {x - 4} \right) - 2\left( {x - 4} \right)}}$
$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)}}{{\left( {x - 4} \right)\left( {x - 2} \right)}}$
$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 1} \right)\left( {{x^2} - 3x - 2x + 6} \right)}}{{\left( {x - 4} \right)\left( {x - 2} \right)}}$
$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 1} \right)\left[ {x\left( {x - 3} \right) - 2\left( {x - 3} \right)} \right]}}{{\left( {x - 4} \right)\left( {x - 2} \right)}}$
$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)\left( {x - 2} \right)}}$
$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)}}$
Now, pass the limit $x = 2$ .
$= \dfrac{{\left( {2 - 1} \right)\left( {2 - 3} \right)}}{{\left( {2 - 4} \right)}}$
$= \dfrac{{1\left( { - 1} \right)}}{{\left( { - 2} \right)}}$
$= \dfrac{1}{2}$
Hence, $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 6x + 8}} = \dfrac{1}{2}$