Given \(\lim\limits_{\text x \to 2}\cfrac{\sqrt{3-\text x}-1}{2-\text x}\)
To find: the limit of the given equation when x tends to 2
Substituting x as 2, we get an indeterminant form of \(\cfrac00\)
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 2
We get \(\lim\limits_{\text x \to 2}\cfrac{\sqrt{3-\text x}-1}{2-\text x}\) =\(\lim\limits_{\text x \to 2}\cfrac{1}{\sqrt{3-\text x}+1}\) = \(\cfrac1{1+1}=\cfrac12\)