Question

How do you find the limit of $\dfrac{{{x^2} - 4}}{{x - 2}}$ as $x$ approaches $2$?

Answer 1

According to the question, first we should expand the term ${x^2} - 4$. We can expand it by using the famous algebraic formula, ${a^2} - {b^2} = (a + b)(a - b)$. We can put the values according to the formula. Then, we can limit the answer with two and get the final answer.

Formula used:
${a^2} - {b^2} = (a + b)(a - b)$

Complete step by step solution:
According to the question, our mathematical expression will be:
$\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{x - 2}}$
Now, we have to expand the term ${x^2} - 4$. To expand this term, we can try to use the algebraic formula which is:
${a^2} - {b^2} = (a + b)(a - b)$
Here, from the formula, we get that $a = x;\,b = 2$
Now, by putting these values according to the formula, we get:
$\Rightarrow {x^2} - 4 = (x + 2)(x - 2)$
Now, when we put the value of ${x^2} - 4$in the question, we get:

Now, we can simplify it. We can cancel the terms. As, we can see that both the numerator and the denominator have the term $x - 2$. So, we can cancel both the terms, and then we will get:
$\Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} (x + 2)$
Now, we will look at the limit part. We will try to solve this part. We will approach $x$as $2$. This means that we need to put the value $2$in place of $x$, then we will get:
$\Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{x - 2}} = (2 + 2)$
$\Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{x - 2}} = 4$
Therefore, we get that the final answer is $4$.

Note:
A limit in mathematics is a value that the sequence(function) is approaching as the index which means that the sequence is approaching some value. These limits are very important to Calculus and these are used to define the integrals, derivatives and continuity.