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Question: How do you evaluate the limit \[\dfrac{x-1}{{{x}^{2}}-1}\] as \[x\] approaches 1?...

How do you evaluate the limit x−1x2−1\dfrac{x-1}{{{x}^{2}}-1} as xx approaches 1?

Explanation

Solution

The above given phrase can be written in a mathematical form as lim⁡x→1x−1x2−1\displaystyle \lim_{x \to 1}\dfrac{x-1}{{{x}^{2}}-1}. We can see in the denominator that the equation is x2−1{{x}^{2}}-1. We can expand this expression using the formula a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=(a-b)(a+b). We will now have the denominator as (x−1)(x+1)(x-1)(x+1). We can see that the numerator and denominator has similar terms so it will get cancelled and then we will apply x→1x \to 1. Hence, we will get the value of the limit.

Complete step by step solution:
According to the given question, we have to solve the equation for xx approaching 1. Firstly, we will begin by writing the mathematical form of the given question, we have,
lim⁡x→1x−1x2−1\displaystyle \lim_{x \to 1}\dfrac{x-1}{{{x}^{2}}-1}-----(1)
We cannot straightaway apply the value of xx as 1, as doing so will cause the expression to have the form 00\dfrac{0}{0}, which is an indeterminate form.
So, we will modify the given expression and after then only we will put xx as 1.
In equation (1), we can see that the denominator has an equation which is x2−1{{x}^{2}}-1 and this can be expanded using an algebraic formula and that is, a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=(a-b)(a+b)
Applying this formula in the denominator, we have the equation (1) as,
⇒lim⁡x→1(x−1)(x−1)(x+1)\Rightarrow \displaystyle \lim_{x \to 1}\dfrac{(x-1)}{(x-1)(x+1)}
We now can see that both the numerator and the denominator have a similar term, (x−1)(x-1) and so this term can be cancelled and the new expression we have is,
⇒lim⁡x→11(x+1)\Rightarrow \displaystyle \lim_{x \to 1}\dfrac{1}{(x+1)}
Now, we will apply the limits x→1x \to 1 in the above expression and we get,
⇒11+1\Rightarrow \dfrac{1}{1+1}
We have,
⇒12\Rightarrow \dfrac{1}{2}
Therefore, lim⁡x→1x−1x2−1=12\displaystyle \lim_{x \to 1}\dfrac{x-1}{{{x}^{2}}-1}=\dfrac{1}{2}.

Note: The expression should be evaluated and reduced in such a way that, when limits are applied to the expression, it should be give the result in the indeterminate forms such as 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty }. Also, certain basic formulae are to be memorized so that solving the expression gets faster.