Question

Evaluate\(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}}\).

Short answer

The value of limit is \(\mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} = \frac{1}{2}\).

Step-by-step solution

Rationalize the given function

Rationalize the given function by multiplying the conjugate of denominator and numerator as follows:

\(\mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} \cdot \frac{{\sqrt {6 - x} + 2}}{{\sqrt {6 - x} + 2}} \cdot \frac{{\sqrt {3 - x} + 1}}{{\sqrt {3 - x} + 1}}\)

Simplify the above expression

Simplify the above obtained expression using formula\(\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\), as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} &= \mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} \cdot \frac{{\sqrt {6 - x} + 2}}{{\sqrt {6 - x} + 2}} \cdot \frac{{\sqrt {3 - x} + 1}}{{\sqrt {3 - x} + 1}}\\ &= \mathop {\lim }\limits_{x \to 2} \,\frac{{{{\left( {\sqrt {6 - x} } \right)}^2} - {2^2}}}{{{{\left( {\sqrt {3 - x} } \right)}^2} - {1^2}}} \cdot \frac{{\sqrt {3 - x} + 1}}{{\sqrt {6 - x} + 2}}\\ &= \mathop {\lim }\limits_{x \to 2} \,\frac{{6 - x - 4}}{{3 - x - 1}} \cdot \frac{{\sqrt {3 - x} + 1}}{{\sqrt {6 - x} + 2}}\\ &= \mathop {\lim }\limits_{x \to 2} \,\frac{{2 - x}}{{2 - x}} \cdot \frac{{\sqrt {3 - x} + 1}}{{\sqrt {6 - x} + 2}}\\\mathop {\lim }\limits_{x \to 2} \,\frac{{\sqrt {3 - x} + 1}}{{\sqrt {6 - x} + 2}}\end{aligned}\).

Apply the limit

Finally, apply the limit by plugging\(x = 2\)into the simplified expression as:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 2} \,\frac{{\sqrt {3 - x} + 1}}{{\sqrt {6 - x} + 2}} &= \frac{{\sqrt {3 - 2} + 1}}{{\sqrt {6 - 2} + 2}}\\ &= \frac{{1 + 1}}{{2 + 2}}\\ &= \frac{2}{4}\\ &= \frac{1}{2}\end{aligned}\)

Thus, the value of limit is \(\mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {6 - x} - 2}}{{\sqrt {3 - x} - 1}} = \frac{1}{2}\).