Question

How do you find the limit of $\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}$ as $x$ approaches 2?

Answer 1

In the given question, we have asked to simplify an expression and the limit of that expression is given. In order to solve the question, first we need to factorize the numerator and denominator one by one to simplify the given expression. As both numerator and denominator have quadratic equations, we will simplify them by splitting the middle term. Later after simplifying we will cancel out the common factor. Then we put x = 2 in the simplified expression. In this way we will get the answer to this question.

Complete step by step answer:
We have given that,
$\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}$
Factorizing the numerator;
We have the numerator as follows,
$3{{x}^{2}}-x-10$
Splitting the middle term of the above equation, we get
$3{{x}^{2}}-6x+5x-10$
Taking out common factor by making the pairs, we get
$3x\left( x-2 \right)+5\left( x-2 \right)$
Taking out the common factors, we get
$\left( 3x+5 \right)\left( x-2 \right)$
Now,
Factorizing the denominator;
We have the denominator as follows,
${{x}^{2}}+5x-14$
Splitting the middle term of the above equation, we get
${{x}^{2}}+7x-2x-14$
Taking out common factor by making the pairs, we get
$x\left( x+7 \right)-2\left( x+7 \right)$
Taking out the common factors, we get
$\left( x+7 \right)\left( x-2 \right)$
Thus,
Combining the numerator and denominator,
We have,
$\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}=\dfrac{\left( 3x+5 \right)\left( x-2 \right)}{\left( x+7 \right)\left( x-2 \right)}$
Cancelling out the common factor, we get
$\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}=\dfrac{\left( 3x+5 \right)}{\left( x+7 \right)}$
Now, putting the limit;
$\underset{x\to 2}{\mathop{\lim }}\,\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}=\underset{x\to 2}{\mathop{\lim }}\,\dfrac{\left( 3x+5 \right)}{\left( x+7 \right)}$
Putting x = 2 in the above fraction, we get
$\underset{x\to 2}{\mathop{\lim }}\,\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}=\underset{x\to 2}{\mathop{\lim }}\,\dfrac{\left( 3x+5 \right)}{\left( x+7 \right)}=\dfrac{\left( 3\left( 2 \right)+5 \right)}{\left( 2+7 \right)}$
Solving the above expression, we get
$\underset{x\to 2}{\mathop{\lim }}\,\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}=\underset{x\to 2}{\mathop{\lim }}\,\dfrac{\left( 3x+5 \right)}{\left( x+7 \right)}=\dfrac{\left( 3\left( 2 \right)+5 \right)}{\left( 2+7 \right)}=\dfrac{11}{9}$
$\therefore\underset{x\to 2}{\mathop{\lim }}\,\dfrac{3{{x}^{2}}-x-10}{{{x}^{2}}+5x-14}=\dfrac{11}{9}$

Hence, $\dfrac{11}{9}$ is the answer.

Note: While solving these types of problems, students need to be very careful while doing the calculation part to avoid making any type of error. They need to know about the concept of the simplification of the quadratic equation. Instead of factoring the polynomial we can solve this question by deriving the numerator and denominator with respect to ‘x’ and then put the given value of ‘x’. Both the ways we will get the same answer.