To solve the limit:

$\lim_{{x \to 1}} (x - 1) \sin\left(2\sqrt{x}\right)$

Let's go step by step:

Step 1: Substituting $x = 1$ directly

When we substitute $x = 1$, the expression becomes:

$(1 - 1) \sin(2\sqrt{1}) = 0 \times \sin(2) = 0$

This gives an indeterminate form $0 \times \sin(2) = 0$. We need a more careful evaluation, such as using L'Hopital's Rule or another method.

Step 2: Using L'Hopital's Rule

L'Hopital's Rule applies to indeterminate forms like $0 \times \infty$ or $\frac{0}{0}$. We can rewrite the expression to apply it:

$\lim_{{x \to 1}} (x - 1) \sin(2\sqrt{x}) = \lim_{{x \to 1}} \frac{\sin(2\sqrt{x})}{\frac{1}{x - 1}}$

We now have the indeterminate form $\frac{0}{0}$, so we can apply L'Hopital's Rule by differentiating the numerator and denominator:

Derivative of the numerator:

$\frac{d}{dx} \left( \sin(2\sqrt{x}) \right) = \cos(2\sqrt{x}) \cdot \frac{d}{dx} (2\sqrt{x}) = \cos(2\sqrt{x}) \cdot \frac{1}{\sqrt{x}}$

Derivative of the denominator:

$\frac{d}{dx} \left( \frac{1}{x - 1} \right) = -\frac{1}{(x - 1)^2}$

Step 3: Applying L'Hopital's Rule

Now, evaluate the limit:

$\lim_{{x \to 1}} \frac{\cos(2\sqrt{x}) \cdot \frac{1}{\sqrt{x}}}{-\frac{1}{(x - 1)^2}}$

This limit should simplify to 0. Therefore, the original limit is:

$\lim_{{x \to 1}} (x - 1) \sin(2\sqrt{x}) = 0$

Let me know if you'd like more details, or if you have any further questions!

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Here are 5 related questions for further practice:

1. What is the limit $\lim_{{x \to 0}} x \sin\left(\frac{1}{x}\right)$?
2. How does L'Hopital's Rule apply to indeterminate forms?
3. What is the derivative of $\sin(3x^2)$?
4. Can you find the limit of $\lim_{{x \to 1}} \frac{x^2 - 1}{x - 1}$?
5. What is the Taylor expansion of $\sin(x)$ around $x = 0$?

Tip: Always check if the limit gives an indeterminate form like $\frac{0}{0}$ or $\infty \times 0$ before applying L'Hopital's Rule.