1. [SOLVED] One-Sided Limit

Evaluate

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

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

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

I split the limit

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

Then I use L'Hopital's Rule

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

which is equivalent as

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

$= ... = \infty$

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.

2. Originally Posted by Paperwings
Evaluate

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

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

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

I split the limit

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

Then I use L'Hopital's Rule

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

which is equivalent as

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

$= ... = \infty$

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.

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

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

I would probably use the reasoning after this step. 1 divided by a number that is positive and getting close to zero gives infinity. t^2 on the other hand, goes to zero. etc.

Not so sure if splitting as a product of two limits (see your work above) is justified. The limit of a product is equal to the product of the limits only if the individual limits exist.

My two cents... Hope this helps.

3. That makes it much easier. Thank you for this:The limit of a product is equal to the product of the limits only if the individual limits exist.