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Question: How do you find the limit of \(\dfrac{{{x^2} - 1}}{{x - 1}}\) as \(x \to 1\)?...

How do you find the limit of x2−1x−1\dfrac{{{x^2} - 1}}{{x - 1}} as x→1x \to 1?

Explanation

Solution

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→1x2−1x−1{\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→1x2−1x−1{\lim _{x \to 1}}\dfrac{{{x^2} - 1}}{{x - 1}}
We know the formula of a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b)
Now we can write our function as lim⁡x→1x2−12x−1{\lim _{x \to 1}}\dfrac{{{x^2} - {1^2}}}{{x - 1}}
Now we will use formula of a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b) then we can write x2−12{x^2} - {1^2} as (x−1)(x+1)(x - 1)(x + 1)
Now we will write our function as lim⁡x→1(x−1)(x+1)(x−1){\lim _{x \to 1}}\dfrac{{(x - 1)(x + 1)}}{{(x - 1)}}
Now after solving we will get the function as lim⁡x→1(x+1){\lim _{x \to 1}}(x + 1)
Now we will put our limit in the function then we will get
=lim⁡x→1(x+1)= {\lim _{x \to 1}}(x + 1)
Now we will put the limit x→1x \to 1 then we will get
=(1+1)= (1 + 1)
=2= 2

Therefore, the solution of our given limit lim⁡x→1x2−1x−1{\lim _{x \to 1}}\dfrac{{{x^2} - 1}}{{x - 1}} is the 22 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 xx so we put 11 at place of xxin the given function after breaking the function.