Question

Find the limit or show that it does not exist.

24. \(\mathop {\lim }\limits_{t \to \infty } \frac{{t + 3}}{{\sqrt {2{t^2} - 1} }}\)

Short answer

The value is \(\mathop {\lim }\limits_{t \to \infty } \frac{{t + 3}}{{\sqrt {2{t^2} - 1} }} = \frac{1}{{\sqrt 2 }}\).

Step-by-step solution

Simply the given limit

Divide the rational function\(\mathop {\lim }\limits_{t \to \infty } \frac{{t + 3}}{{\sqrt {2{t^2} - 1} }}\)in the given limit by\(t\)and simplify as shown below:

\(\mathop {\lim }\limits_{t \to \infty } \frac{{t + 3}}{{\sqrt {2{t^2} - 1} }} = \mathop {\lim }\limits_{t \to \infty } \frac{{\left( {t + 3} \right)/t}}{{\sqrt {2{t^2} - 1} /t}}\)

Simply further

Perform the division in numerator and denominator and apply necessary limit laws.

As \(t = \sqrt {{t^2}} \), for \(t > 0\), division can be performed in the square root positively.

So, simplify further as shown below:

\(\begin{aligned}{c}\mathop {\lim }\limits_{t \to \infty } \frac{{t + 3}}{{\sqrt {2{t^2} - 1} }} &= \mathop {\lim }\limits_{t \to \infty } \frac{{\left( {t + 3} \right)/t}}{{\sqrt {2{t^2} - 1} /t}}\\ &= \mathop {\lim }\limits_{t \to \infty } \frac{{\left( {t + 3} \right)/t}}{{\sqrt {2{t^2} - 1/{t^2}} }}\\ &= \mathop {\lim }\limits_{t \to \infty } \frac{{1 + 3/t}}{{\sqrt {2 - 1/{t^2}} }}\\ &= \frac{{\mathop {\lim }\limits_{t \to \infty } \left( {1 + 3/t} \right)}}{{\sqrt {\mathop {\lim }\limits_{t \to \infty } \left( {2 - 1/{t^2}} \right)} }}\\ &= \frac{{\mathop {\lim }\limits_{t \to \infty } 1 + \mathop {\lim }\limits_{t \to \infty } \left( {3/t} \right)}}{{\sqrt {\mathop {\lim }\limits_{t \to \infty } 2 - \mathop {\lim }\limits_{t \to \infty } \left( {1/{t^2}} \right)} }}\end{aligned}\).

Finally, apply the limit 

On applying the limit, to the above equation, and simplifying we get:

\(\begin{aligned}{c}\mathop {\lim }\limits_{t \to \infty } \frac{{t + 3}}{{\sqrt {2{t^2} - 1} }} &= \frac{{\mathop {\lim }\limits_{t \to \infty } 1 + \mathop {\lim }\limits_{t \to \infty } \left( {3/t} \right)}}{{\sqrt {\mathop {\lim }\limits_{t \to \infty } 2 - \mathop {\lim }\limits_{t \to \infty } \left( {1/{t^2}} \right)} }}\\ &= \frac{{1 + 0}}{{\sqrt {2 - 0} }}\\ &= \frac{1}{{\sqrt 2 }}\end{aligned}\)

Thus, \(\mathop {\lim }\limits_{t \to \infty } \frac{{t + 3}}{{\sqrt {2{t^2} - 1} }} = \frac{1}{{\sqrt 2 }}\).