Question

What is the limit of $\left( \dfrac{1}{x-1}-\dfrac{2}{{{x}^{2}}-1} \right)$ as x approaches $1$ ?

Answer 1

Hint: If we see the given function, we can see that if we directly put $x=1$ in the given function, it becomes undefined. Also, the form is not that of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ , so L’ Hospital rule can also not be applied. The only way is to factorise ${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$ and then simplify the function to $\left( \dfrac{1}{x-1}-\dfrac{2}{\left( x+1 \right)\left( x-1 \right)} \right)$ and then finally to $\left( \dfrac{1}{x+1} \right)$ . Applying the limits, we get $\dfrac{1}{2}$.

Complete step by step solution:
The given function that we have in the given problem of limits is,
$\left( \dfrac{1}{x-1}-\dfrac{2}{{{x}^{2}}-1} \right)$
Now, we know that ${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$ . The second term of the above given function can thus be written as $\dfrac{2}{\left( x+1 \right)\left( x-1 \right)}$ and the entire above function can thus be written as,
$\Rightarrow \left( \dfrac{1}{x-1}-\dfrac{2}{\left( x+1 \right)\left( x-1 \right)} \right)$
Now, we have two fractions. One of the two fractions is $\dfrac{1}{x-1}$ and the other one of the two fractions is $\dfrac{2}{\left( x+1 \right)\left( x-1 \right)}$ . In order to subtract the second fraction from the first fraction, we take the LCM of the denominators. The denominator will be nothing but $\left( x+1 \right)\left( x-1 \right)$ . The equivalent fraction of $\dfrac{1}{x-1}$ with denominator $\left( x+1 \right)\left( x-1 \right)$ is $\dfrac{x+1}{\left( x+1 \right)\left( x-1 \right)}$. Thus, the subtraction of the two above fractions after taking the common denominator of the two will become,
 $\Rightarrow \left( \dfrac{x+1}{\left( x+1 \right)\left( x-1 \right)}-\dfrac{2}{\left( x+1 \right)\left( x-1 \right)} \right)$
Carrying on the subtraction, the function becomes,
$\Rightarrow \left( \dfrac{x+1-2}{\left( x+1 \right)\left( x-1 \right)} \right)$
Upon simplification, the above function becomes,
$\Rightarrow \left( \dfrac{x-1}{\left( x+1 \right)\left( x-1 \right)} \right)$
Further simplification of the above formula leads us to,
$\Rightarrow \left( \dfrac{1}{x+1} \right)$
Taking the limits as x tends to approaches $1$ , the function becomes,
\[\Rightarrow \displaystyle \lim_{x \to 1}\left( \dfrac{1}{x+1} \right)\]
Putting $x=1$ in the above limits, we get,
\[\Rightarrow \displaystyle \lim_{x \to 1}\left( \dfrac{1}{x+1} \right)=\dfrac{1}{2}\]
Thus, we can conclude that the limit of $\left( \dfrac{1}{x-1}-\dfrac{2}{{{x}^{2}}-1} \right)$ as x approaches $1$ will be $\dfrac{1}{2}$.

Note: The first and one of the most common mistakes that students make is directly putting $x=1$ . This gives the result as infinity. But this is not the correct answer. We should have patience and slowly carry out the subtraction of fractions and turn it into a simpler form and then apply the limits.