Key Concepts and Formulas

• Limits and Indeterminate Forms: Understanding when a limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, which suggests the need for further evaluation.
• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Chain Rule for Differentiation: If $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Step-by-Step Solution

Step 1: Evaluate the limit and identify the indeterminate form. We are asked to find the limit: $L = \mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}}$ Let's substitute $x=1$ into the expression. We are given that $f(1) = 1$. Numerator: $\sqrt{f(1)} - 1 = \sqrt{1} - 1 = 1 - 1 = 0$. Denominator: $\sqrt{1} - 1 = 1 - 1 = 0$. Since we get the form $\frac{0}{0}$, this is an indeterminate form, and we can consider using L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule. L'Hôpital's Rule states that if $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$. Let the numerator function be $N(x) = \sqrt{f(x)} - 1$ and the denominator function be $D(x) = \sqrt{x} - 1$. We need to find the derivatives of $N(x)$ and $D(x)$ with respect to $x$.

Using the chain rule for the numerator $N(x)$: $\frac{d}{dx}(\sqrt{f(x)} - 1) = \frac{d}{dx}(f(x)^{1/2}) - \frac{d}{dx}(1)$ $= \frac{1}{2} f(x)^{-1/2} \cdot f'(x) - 0$ $= \frac{f'(x)}{2\sqrt{f(x)}}$

Using the power rule for the denominator $D(x)$: $\frac{d}{dx}(\sqrt{x} - 1) = \frac{d}{dx}(x^{1/2}) - \frac{d}{dx}(1)$ $= \frac{1}{2} x^{-1/2} - 0$ $= \frac{1}{2\sqrt{x}}$

Now, applying L'Hôpital's Rule: $L = \mathop {\lim }\limits_{x \to 1} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$

Step 3: Simplify and evaluate the new limit. We can simplify the expression by multiplying the numerator by the reciprocal of the denominator: $L = \mathop {\lim }\limits_{x \to 1} \frac{f'(x)}{2\sqrt{f(x)}} \cdot \frac{2\sqrt{x}}{1}$ $L = \mathop {\lim }\limits_{x \to 1} \frac{f'(x)\sqrt{x}}{\sqrt{f(x)}}$ Now, substitute $x=1$ into this simplified expression, using the given values $f(1)=1$ and $f'(1)=2$: $L = \frac{f'(1)\sqrt{1}}{\sqrt{f(1)}}$ $L = \frac{(2)(1)}{\sqrt{1}}$ $L = \frac{2}{1}$ $L = 2$

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule.
• Errors in Differentiation: Carefully apply the chain rule and power rule when differentiating composite functions like $\sqrt{f(x)}$. A common mistake is forgetting to multiply by $f'(x)$ when differentiating $\sqrt{f(x)}$.
• Algebraic Simplification: Incorrectly simplifying the fraction after applying L'Hôpital's Rule can lead to errors in the final calculation.

Summary

The problem asks for the evaluation of a limit that results in an indeterminate form $\frac{0}{0}$. We successfully applied L'Hôpital's Rule by differentiating the numerator and the denominator separately. After differentiating, we obtained a new limit expression. Substituting the given values of $f(1)$ and $f'(1)$ into this new expression allowed us to directly calculate the value of the limit, which turned out to be 2.

The final answer is \boxed{2}.