The given limit problem is:

$\lim_{{x \to 1^+}} \left( \frac{1}{\sqrt{x-1}} - \frac{1}{\sqrt{x^2-1}} \right)$

Let's analyze the limit step by step.

1. Simplification of the second term: The expression inside the second square root is $x^2 - 1$, which can be factored: $x^2 - 1 = (x - 1)(x + 1)$ Thus, the second term becomes: $\frac{1}{\sqrt{x^2 - 1}} = \frac{1}{\sqrt{(x-1)(x+1)}} = \frac{1}{\sqrt{x-1} \cdot \sqrt{x+1}}$

2. Rewrite the original expression: Substituting this into the original limit, we have: $\lim_{{x \to 1^+}} \left( \frac{1}{\sqrt{x-1}} - \frac{1}{\sqrt{x-1} \cdot \sqrt{x+1}} \right)$

3. Factor out $\frac{1}{\sqrt{x-1}}$: $\lim_{{x \to 1^+}} \frac{1}{\sqrt{x-1}} \left( 1 - \frac{1}{\sqrt{x+1}} \right)$

4. Examine the limit as $x \to 1^+$:

   • As $x \to 1^+$, $\sqrt{x - 1} \to 0^+$.
   • $\sqrt{x+1} \to \sqrt{2}$.

   So, the expression becomes: $\lim_{{x \to 1^+}} \frac{1}{0^+} \left( 1 - \frac{1}{\sqrt{2}} \right)$ The term inside the parentheses approaches a constant value: $1 - \frac{1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2} \approx 1 - 0.7071 = 0.2929$

5. Conclusion: The expression simplifies to: $\frac{1}{0^+} \cdot 0.2929$ Since $\frac{1}{0^+}$ tends to infinity, the overall limit tends to infinity. Therefore:

$\boxed{\infty}$

The limit as $x$ approaches 1 from the right is infinity.

Would you like more details on any step or have further questions?

Here are some related questions:

1. How do you factor and simplify expressions like $x^2 - 1$?
2. What is the meaning of $x \to 1^+$ in limit problems?
3. How does multiplying by $\frac{1}{0^+}$ lead to infinity?
4. How do square roots affect limit behavior near boundary points?
5. What strategies can simplify expressions involving square roots?

Tip: In limits involving indeterminate forms or complex expressions, factoring and cancellation often help to simplify the problem!