Question

Evaluate the limit, if it exists. (if an answer does not exist, enter dne. ) lim t→0 1 t 1 t − 1 t.

Answer 1

The result of the limit is $\infty$.

What is the limit?

The limit that we want to find is:

$\lim_{t \rightarrow 0} \frac{1}{t\sqrt{1 + t}} - \frac{1}{t}$

Using the common factor, we have that:

$\lim_{t \rightarrow 0} \frac{1}{t\sqrt{1 + t}} - \frac{1}{t} = \lim_{t \rightarrow 0} \frac{1}{t} \left(\frac{1}{\sqrt{1+t}} - 1\right)$

Rationalizing the expression inside the parenthesis:

$\left(\frac{1}{\sqrt{1+t} - 1}\right) \times \frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1} = \frac{\sqrt{1+t} + 1}{t}$

Then, the limit is:

$\lim_{t \rightarrow 0} \frac{1}{t} \left(\frac{1}{\sqrt{1+t} - 1}\right) = \lim_{t \rightarrow 0} \frac{1}{t^2} \sqrt{1+t} + 1$

Applying lateral limits:

$\lim_{t \rightarrow 0^-} \frac{1}{t^2} \sqrt{1+t} + 1 = \infty$

$\lim_{t \rightarrow 0^+} \frac{1}{t^2} \sqrt{1+t} + 1 = \infty$

Since the lateral limits are equal, the limit goes to infinity.

You can learn more about limits at https://brainly.com/question/26270080