Question

Find the value of \[\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right)\]

Answer 1

Hint: Here, we will solve the quadratic equations present in the denominators using middle term splitting. We will simplify the expression further by taking the LCM and canceling out the like terms. We will then apply the limit by substituting the value of \[x = 3\] and simplify further to find the required value of the given limit.

Complete step-by-step answer:
As we can see, there are quadratic equations in the denominators hence, we will do the middle term split.
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 8x - 3x + 24}} + \dfrac{1}{{{x^2} - 3x + 2x - 6}}} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{x\left( {x - 8} \right) - 3\left( {x - 8} \right)}} + \dfrac{1}{{x\left( {x - 3} \right) + 2\left( {x - 3} \right)}}} \right)\]
Factoring out the common terms, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{\left( {x - 3} \right)\left( {x - 8} \right)}} + \dfrac{1}{{\left( {x + 2} \right)\left( {x - 3} \right)}}} \right)\]
Now taking \[\dfrac{1}{{x - 3}}\] common from both the fractions, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \dfrac{1}{{\left( {x - 3} \right)}}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{\left( {x - 8} \right)}} + \dfrac{1}{{\left( {x + 2} \right)}}} \right)\]
Now, taking the LCM inside the bracket, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \dfrac{1}{{\left( {x - 3} \right)}}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{x + 2 + x - 8}}{{\left( {x - 8} \right)\left( {x + 2} \right)}}} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \dfrac{1}{{\left( {x - 3} \right)}}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{2x - 6}}{{\left( {x - 8} \right)\left( {x + 2} \right)}}} \right)\]
Taking 2 common from the numerator, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \dfrac{2}{{\left( {x - 3} \right)}}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{\left( {x - 3} \right)}}{{\left( {x - 8} \right)\left( {x + 2} \right)}}} \right)\]
Cancelling out \[\left( {x - 3} \right)\] from the numerator as well as the denominator, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = 2\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{\left( {x - 8} \right)\left( {x + 2} \right)}}} \right)\]
Now, removing the limit by substituting \[x = 3\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = 2\left( {\dfrac{1}{{\left( {3 - 8} \right)\left( {3 + 2} \right)}}} \right)\]
Adding and subtracting the terms in the denominator, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \left( {\dfrac{2}{{\left( { - 5} \right)\left( 5 \right)}}} \right) = - \dfrac{2}{{25}}\]
Therefore, the value of \[\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \dfrac{{ - 2}}{{25}}\]
Hence, this is the required answer.

Note: In mathematics, a limit is a value that a function approaches as the input or the index approaches some value. Limits are an essential element of calculus and are used to define continuity, integrals and derivatives. We usually put the limit at the end and not in the starting because it gives an answer as 0 or infinity.
Now let’s see that if we substitute \[x = 3\] and hence remove the limit, what will be our value of the given limit.
\[\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \left( {\dfrac{1}{{{3^2} - 11\left( 3 \right) + 24}} + \dfrac{1}{{{3^2} - 3 - 6}}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{1}{{{x^2} - 11x + 24}} + \dfrac{1}{{{x^2} - x - 6}}} \right) = \left( {\dfrac{1}{{9 - 33 + 24}} + \dfrac{1}{{9 - 9}}} \right) = \dfrac{1}{0}\]
We can see that there is a 0 in the denominator hence, we will have to first simplify the equation and then substitute the value of \[x\].