Question

Given $\dfrac{{{x}^{2}}-9}{x-3}$, how do you find the limit as x approaches 3?

Answer 1

Since, in the question the limit given to us is making the function tending to not defined because of the denominator as 0 so, we will use the algebraic formula $a^2-b^2=(a+b)(a-b)$ and solve the function to get the desired limit. We will try to eliminate the denominator in order to get the answer. And this can be done by using the algebraic formula.

Complete step-by-step answer:
To get to know the closeness between the function and its limit we will use a quadratic algebraic equation. For this we will consider the given function $\dfrac{{{x}^{2}}-9}{x-3}$ as x approaches the point 3. This formula is ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We can consider ${{x}^{2}}-{{3}^{2}}$ instead of ${{x}^{2}}-9$. Therefore, we get ${{x}^{2}}-{{3}^{2}}=\left( x+3 \right)\left( x-3 \right)\,\,\,\,...(i)$.
We will start by considering the given function $\displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)$. In this function we will use equation (i). Therefore, we get

& \displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)=\displaystyle \lim_{x \to 3}\dfrac{{{x}^{2}}-{{3}^{2}}}{x-3} \\\ & \Rightarrow \displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)=\displaystyle \lim_{x \to 3}\dfrac{\left( x-3 \right)\left( x+3 \right)}{\left( x-3 \right)} \\\ \end{aligned}$$ Now, we will cancel the common term, that is (x – 3), from the numerator and denominator. Thus we get, $$\begin{aligned} & \displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)=\displaystyle \lim_{x \to 3}\dfrac{\left( x-3 \right)\left( x+3 \right)}{\left( x-3 \right)} \\\ & \Rightarrow \displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)=\displaystyle \lim_{x \to 3}\left( x+3 \right) \\\ \end{aligned}$$ This is the point at which we need to substitute the value of x in the above equation. This will result into, $$\begin{aligned} & \displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)=3+3 \\\ & \Rightarrow \displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)=6 \\\ \end{aligned}$$ Hence, the value of the limit $$\displaystyle \lim_{x \to 3}\left( \dfrac{{{x}^{2}}-9}{x-3} \right)=6$$. **Note:** In order to solve such type of questions, we need to take care of the denominator. If after substituting the value of the limit we are getting 0, then the only answer we will get is going to be not defined. So, to reduce such problems, we should not substitute the limit into the function at first. Instead of this we will use some algebraic formula to solve it. Here, the quadratic formula that we used here is ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We should note that there is a difference between these two formulas, ${{\left( a-b \right)}^{2}}$ and ${{a}^{2}}-{{b}^{2}}$. If any wrong formula is used then, instead of knowing the correct we will not be able to get the desired answer.