Question

Find the limit or show that it does not exist.

31. \(\mathop {lim}\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} } \right)\)

Short answer

The value of the limit is \(\frac{{a - b}}{2}\)

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:

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

Thus, the value of the limit is \(\frac{{a - b}}{2}\).