Question

Find the limit or show that it does not exist.

32. \(\mathop {lim}\limits_{x \to \infty } \left( {x - \sqrt x } \right)\)

Short answer

The value of the limit is \(\infty \).

Step-by-step solution

Use the property of limit

A function is continuous at a point “a” if \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\).

Evaluate the given limit

Evaluate the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } \left( {x - \sqrt x } \right){\rm{ }} &= \mathop {\lim }\limits_{x \to \infty } \left( {x - \sqrt x } \right) \times \frac{{\left( {x + \sqrt x } \right)}}{{{{(x + \sqrt x )}^2}}}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( x \right)}^2} - {{\left( {\sqrt x } \right)}^2}}}{{x\left( {1 + \frac{1}{{\sqrt x }}} \right)}}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{x^2} - x} \right)}}{{x\left( {1 + \frac{1}{{\sqrt x }}} \right)}}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{x^2} - 1} \right)}}{{x\left( {1 + \frac{1}{{\sqrt x }}} \right)}}\\ &= \frac{{x\left( {x - 1} \right)}}{{x\left( {1 + \frac{1}{{\sqrt x }}} \right)}}\\ &= \left( {\frac{{\infty - 1}}{{1 + \frac{1}{{\sqrt x }}}}} \right)\\ &= \left( {\frac{\infty }{{1 + 0}}} \right)\\ &= \infty \end{aligned}\)

Thus, the value of the limit is \(\infty \).