Find a number a such that the following limit exits. \lim\limits_{x\to 3} \left(...


Find a number {eq}a {/eq} such that the following limit exits.

{eq}\lim\limits_{x\to 3} \left( \frac{2x^2-3ax+x-a-1}{x^2-2x-3} \right) {/eq}

Evaluate the limit.


Here, we have to find the value of constant such the limit of the given function exists.

The limit of a function exists only if the denominator of the function at a given point is not equal to zero.

By direct substitution, we find that the denominator turns out to be 0. So, first, we factor the denominator and then try to cancel the factor so that the denominator will no longer be 0 and the limit will exist.

Answer and Explanation:

We have to find a number {eq}a {/eq} such that the following limit exists.

{eq}\lim\limits_{x\to 3} \left( \dfrac{2x^2-3ax+x-a-1}{x^2-2x-3} \right) {/eq}.

Since the limit exists, the denominator must not equal to zero.


{eq}\begin{align} x^2-2x-3 &=x^2-3x+x-3\\ &=x(x-3)+1(x-3)\\ &=(x+1)(x-3) \end{align} {/eq}

So the denominator contains exactly one factor {eq}(x-3) {/eq} at which the limit does not exist.

So in order that {eq}\lim\limits_{x\to 3} \left( \dfrac{2x^2-3ax+x-a-1}{x^2-2x-3} \right) {/eq} has a limit the only requirement is that:

{eq}2x^2-3ax+x-a-1 {/eq} is divisible by {eq}(x-3) {/eq}

Let {eq}f(x)=2x^2-3ax+x-a-1 {/eq}

This is divisible by {eq}(x-3) {/eq} iff {eq}f(3)=0 {/eq}

Substituting {eq}x=3 {/eq}, we have:

{eq}\begin{align} f(3) &=2(3)^2-3a(3)+(3)-a-1\\ 0 &=18-9a+3-a-1\\ 0 &=20-10a\\ 20 &=10a\\ a &=2 \end{align} {/eq}

With this value of {eq}a {/eq}

{eq}\begin{align} \lim\limits_{x\to 3} \left( \dfrac{2x^2-3ax+x-a-1}{x^2-2x-3} \right) &=\lim\limits_{x\to 3} \left( \dfrac{2x^2-3(2)x+x-2-1}{x^2-2x-3} \right)\\ &=\lim\limits_{x\to 3} \left( \dfrac{2x^2-5x-3}{x^2-2x-3} \right)\\ &=\lim\limits_{x\to 3} \left( \dfrac{2x^2-6x+x-3}{(x-3)(x+1)} \right)\\ &=\lim\limits_{x\to 3} \left( \dfrac{2x(x-3)+(x-3)}{(x-3)(x+1)} \right)\\ &=\lim\limits_{x\to 3} \left( \dfrac{(2x+1)(x-3)}{(x-3)(x+1)} \right)\\ &=\lim\limits_{x\to 3} \left( \dfrac{2x+1}{(x+1)} \right)\\ &=\dfrac{2(3)+1}{3+1}\\ &=\dfrac{7}{4} \end{align} {/eq}

{eq}\color{blue}{a=2 \: \text{and the limit is} \: \dfrac{7}{4}} {/eq}

Learn more about this topic:

How to Determine if a Limit Does Not Exist

from AP Calculus AB: Exam Prep

Chapter 4 / Lesson 9

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