Question

Find the limit or show that it does not exist.

29.\(\mathop {lim}\limits_{t \to \infty } \left( {\sqrt {25{t^2} + 2} + - 5t} \right)\)

Short answer

The value of the limit is 0.

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:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {25{t^2} + 2} - 5t} \right){\rm{ }} &= \mathop {\lim }\limits_{t \to 0} {\left( {\sqrt {25{t^2} + 2} - 5t} \right)^2} \times \frac{{\left( {\sqrt {25{t^2} + 2} + 5t} \right)}}{{\left( {\sqrt {25{t^2} + 2} + 5t} \right)}}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {\sqrt {25{t^2} + 2} } \right)}^2} - {{\left( {5t} \right)}^2}}}{{\left( {\sqrt {{t^2}\left( {25 + \frac{2}{{{t^2}}}} \right)} + 5} \right)}}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{25{t^2} + 2 - 25{t^2}}}{{t\left( {\sqrt {25 + \frac{2}{{{t^2}}}} + 5} \right)}}\\ &= \frac{2}{{\infty \left( {\sqrt {25 + \frac{2}{{{\infty ^2}}}} + 5} \right)}}\\ &= \frac{2}{5}\\ &= 0\end{aligned}\)

Thus, the value of the limit is 0.