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**Paperwings** Evaluate

$\displaystyle \lim_{t\to0^{+}} \frac{\sqrt{1+t^3}}{\sqrt{t}} $

I multiply the numerator and denominator by $\displaystyle 1 = \frac{\sqrt{t}}{\sqrt{t}} $ to get

$\displaystyle \lim_{t\to0^{+}} \frac{\sqrt{t} \sqrt{t+t^3} }{t} $

I split the limit

$\displaystyle \lim_{t\to0^{+}} \frac{\sqrt{t}}{t} \lim_{t\to0^{+}} \frac{\sqrt{t+t^3} }{t} $

Then I use L'Hopital's Rule

$\displaystyle \lim_{t\to0^{+}} \frac{1}{2}t^{-1/2} \lim_{t\to0^{+}} \frac{1}{2}(1+t^3)^{-1/2}\cdot 3t^2 $

which is equivalent as

$\displaystyle \lim_{t\to0^{+}} \frac{1}{2\sqrt{t}} \lim_{t\to0^{+}} \frac{3t^2}{2\sqrt{1+t^3}} $

$\displaystyle = ... = \infty $

My reasoning is since the denominator the left limit gets larger and larger approaching infinity and the limit of the right is always greater than 0, the entire limit is infinity.