Question

How do you find the limit of $\dfrac{{{x^2} - 1}}{{x - 1}}$ as $x \to 1$?

Answer 1

Hint: When we discuss the limits then always remember that we have to break a given function in the form of a function where when we put the limit in the function then we get a finite value not an infinite value.

Complete step by step Solution:
Given that –
${\lim _{x \to 1}}\dfrac{{{x^2} - 1}}{{x - 1}}$
Now in above function if we will put the limit then we will get the infinite value of function
Now first we will solve the function then we try to put our limit so we will get a finite value
Now we will solve our function ${\lim _{x \to 1}}\dfrac{{{x^2} - 1}}{{x - 1}}$
We know the formula of ${a^2} - {b^2} = (a - b)(a + b)$
Now we can write our function as ${\lim _{x \to 1}}\dfrac{{{x^2} - {1^2}}}{{x - 1}}$
Now we will use formula of ${a^2} - {b^2} = (a - b)(a + b)$ then we can write ${x^2} - {1^2}$ as $(x - 1)(x + 1)$
Now we will write our function as ${\lim _{x \to 1}}\dfrac{{(x - 1)(x + 1)}}{{(x - 1)}}$
Now after solving we will get the function as ${\lim _{x \to 1}}(x + 1)$
Now we will put our limit in the function then we will get
$ = {\lim _{x \to 1}}(x + 1)$
Now we will put the limit $x \to 1$ then we will get
$ = (1 + 1)$
$ = 2$

Therefore, the solution of our given limit ${\lim _{x \to 1}}\dfrac{{{x^2} - 1}}{{x - 1}}$ is the $2$ which is our required answer for the our question.

Note:
Always remember the value after putting the limit in the function is always a finite value and when we put a limit then we will put the value of variable for example in above question our variable is $x$ so we put $1$ at place of $x$in the given function after breaking the function.