The given expression is:

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

Step-by-Step Solution:

1. Identify the form of the expression: As $t \to 0$, the denominator becomes $0 \cdot \infty$ or indeterminate because the term $t \sqrt{1+t} - t$ tends to $0$ as $t \to 0$.

2. Simplifying the denominator: The denominator is $t(\sqrt{1+t} - 1)$. This form can be simplified using a trick involving multiplying and dividing by the conjugate of $\sqrt{1+t} - 1$.

   Multiply both the numerator and denominator by $\sqrt{1+t} + 1$:

   $\lim_{t \to 0} \frac{1}{t(\sqrt{1+t} - 1)} \cdot \frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1}$

3. Simplification: The denominator becomes:

   $t \cdot \frac{(\sqrt{1+t} - 1)(\sqrt{1+t} + 1)}{\sqrt{1+t} + 1} = t \cdot \frac{1+t - 1}{\sqrt{1+t} + 1} = t \cdot \frac{t}{\sqrt{1+t} + 1}$

   So, the expression simplifies to:

   $\lim_{t \to 0} \frac{\sqrt{1+t} + 1}{t^2}$

4. Taking the limit: As $t \to 0$, $\sqrt{1+t} \to 1$. Therefore, the expression becomes:

   $\lim_{t \to 0} \frac{1 + 1}{t^2} = \lim_{t \to 0} \frac{2}{t^2}$

   Since $\frac{2}{t^2} \to \infty$ as $t \to 0$, the limit diverges to infinity.

Final Answer:

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

Would you like more details on any part of the solution, or do you have additional questions? Here are some related questions you might find useful:

1. How does multiplying by the conjugate help in simplifying limits involving square roots?
2. What other indeterminate forms can be handled using similar techniques?
3. Can this limit be solved using L'Hopital's Rule instead?
4. Why does the expression tend toward infinity as $t \to 0$?
5. How can this type of problem be related to real-world applications of limits?

Tip: When simplifying expressions with square roots, the conjugate is often a helpful tool to remove irrational terms and make limits easier to compute.