Aspire Faculty ID #11550 · Topic: NIMCET 2023 · Just now
NIMCET 2023

Let $f(x)=\frac{x^2-1}{|x|-1}$. Then the value of $lim_{x\to-1} f(x)$ is

Solution

Given $f(x)=\dfrac{x^2-1}{|x|-1}$ 
$x=-2$
So $f(x)=\dfrac{(x-1)(x+1)}{-(x+1)}$ Cancel $(x+1)$: $f(x)=-(x-1)=1-x$ 
Now take the limit: 
$\lim_{x\to -1}(1-x)$
$=1-(-1)=2$

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