Question

How do you evaluate the limit of $ \lim \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} $ as $ x \to 3 $ ?

Answer 1

Hint: In order to evaluate the limit of $ \lim \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} $ as $ x \to 3 $ put $ x \to 3 $ in the equation and check whether it is in the form of indeterminate form or not i.e $ \dfrac{0}{0},\dfrac{\infty }{\infty } $ ,etc. If yes then reduce the equation to its simplest form and when it reaches the stage where no more simplification can be then just apply $ x \to 3 $ as value and our limit is obtained.

Complete step by step solution:
We are given with the equation $ \lim \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} $ as $ x \to 3 $ which be written as $ \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} $ .
To check whether it is in indeterminate form or not just $ x \to 3 $ in $ \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} $ and we get:
 $ \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{{3^2} - 3 \times 3}}{{{3^2} + 2 \times 3 - 15}} = \dfrac{{9 - 9}}{{9 + 6 - 15}} = \dfrac{0}{0} $
And the result obtained is in the indeterminate form so simplify it to the lowest form:
For simplifying take $ x $ common from numerator and simplify the denominator using mid term factorization and we get:
 $
  \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} \\
   = \mathop {\lim }\limits_{x \to 3} \dfrac{{x\left( {x - 3} \right)}}{{{x^2} + 5x - 3x - 15}} \\
   = \mathop {\lim }\limits_{x \to 3} \dfrac{{x\left( {x - 3} \right)}}{{x\left( {x + 5} \right) - 3(x + 5)}} \\
  \mathop {\lim }\limits_{x \to 3} \dfrac{{x\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 5} \right)}} \;
  $
We can cancel the same terms from numerator and denominator and we get:
 $
  \mathop {\lim }\limits_{x \to 3} \dfrac{{x\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 5} \right)}} \\
   = \mathop {\lim }\limits_{x \to 3} \dfrac{x}{{\left( {x + 5} \right)}} \\
  $
Since, we can see that it cannot be further simplified so just put the value of $ x \to 3 $ , then we get:
\[
  \mathop {\lim }\limits_{x \to 3} \dfrac{x}{{\left( {x + 5} \right)}} \\
   = \dfrac{{\mathop {\lim }\limits_{x \to 3} x}}{{\mathop {\lim }\limits_{x \to 3} \left( {x + 5} \right)}} \\
   = \dfrac{3}{{3 + 5}} = \dfrac{3}{8} \;
 \]
Therefore, the limit of $ \lim \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 15}} $ as $ x \to 3 $ is $ \dfrac{3}{8} $ .
So, the correct answer is “ $ \dfrac{3}{8} $ ”.

Note: We are given the Quadratic equation in the middle so, we have solved the denominator using mid-term factorization which is splitting up the middle and then taking common and then cancelling the common values.
But, since it was a Quadratic equation so, we can solve that using Quadratic formula also, which would also give the same value.