If f(1) = 1, f'(1) = 2, then lim (x-1) (sqrt(f(x)) - 1) / (sqrt(x) - 1) is (A) 2 (B) 4 (C) 1 (D) 1/

Q: Why can't we just plug in $x=1$ immediately?

If you plug in $x=1$ directly, you get $\frac{\sqrt{f(1)} - 1}{\sqrt{1} - 1} = \frac{\sqrt{1} - 1}{1 - 1} = \frac{0}{0}$. This is an indeterminate form. It doesn't mean the limit doesn't exist; it means the function is hiding its true behavior at that point. We need to use tools like L'Hopital's Rule to 'unmask' the value.

Q: Why did we use the Chain Rule for $\sqrt{f(x)}$?

Great question! $\sqrt{f(x)}$ is a composite function—it's the square root function applied to $f(x)$. The Chain Rule tells us that the derivative of $g(f(x))$ is $g'(f(x)) \cdot f'(x)$. Here, $g(u) = \sqrt{u}$, so $g'(u) = \frac{1}{2\sqrt{u}}$. Thus, the derivative is $\frac{1}{2\sqrt{f(x)}} \cdot f'(x)$.

Q: Is L'Hopital's Rule the only way to solve this?

Not at all! You could also solve this using the definition of the derivative. Since $\lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = f'(1)$, you can manipulate the original expression to resemble this definition. However, L'Hopital's Rule is often faster and more direct for these types of problems.

Question

If $f(1) = 1, f'(1) = 2$, then $\lim_{x 	o 1} \frac{\sqrt{f(x)} - 1}{\sqrt{x} - 1}$ is

Options:

(A) 2

(B) 4

(C) 1

(D) 1/2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the limit:

$$\lim_{x 	o 1} \frac{\sqrt{f(x)} - 1}{\sqrt{x} - 1}$$

We are provided with two vital clues: $f(1) = 1$ and $f'(1) = 2$. These values serve as the essential keys to unlocking the solution.

### The Indeterminate Trap

Whenever you face a limit, your first instinct should always be direct substitution. If we plug $x=1$ into our expression, the numerator becomes $\sqrt{f(1)} - 1 = \sqrt{1} - 1 = 0$.

The denominator similarly becomes $\sqrt{1} - 1 = 0$. We have arrived at the indeterminate form $\frac{0}{0}$.

In the world of calculus, this is not a failure; it is a gateway. It is a sign that the function is behaving in a way that requires further investigation. To resolve this, we apply **L'Hopital's Rule**.

### The Power of L'Hopital's Rule

L'Hopital's Rule states that for a limit of the form $\frac{0}{0}$, we can differentiate the numerator and the denominator separately to find the limit.

For the numerator, we apply the Chain Rule to $\frac{d}{dx}(\sqrt{f(x)} - 1)$:

$$\frac{d}{dx}(\sqrt{f(x)}) = \frac{f'(x)}{2\sqrt{f(x)}}$$

For the denominator, the derivative of $\sqrt{x} - 1$ is a standard result:

$$\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$$

### The Grand Simplification

Now, we reconstruct our limit using these derivatives:

$$\lim_{x 	o 1} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$$

Notice that the factor of $\frac{1}{2}$ appears in both the numerator and the denominator, allowing them to cancel out. We are left with the simplified expression:

$$\lim_{x 	o 1} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$$

### The Final Reveal

The indeterminate form has now vanished. We can safely substitute $x=1$ into the expression:

$$\frac{f'(1) \cdot \sqrt{1}}{\sqrt{f(1)}}$$

Using our original clues $f'(1) = 2$ and $f(1) = 1$, we calculate:

$$\frac{2 \cdot 1}{\sqrt{1}} = \frac{2}{1} = 2$$

The final value of the limit is **2**. This result demonstrates how the systematic application of calculus tools can resolve complex-looking indeterminate forms into elegant solutions.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (1) = 1, f^{'} (1) = 2$ , then $lim_{x \to 1} \frac{f ( x ) - 1}{x - 1}$ is

Select Answer:

Visualized Solution

Analyze Given Values

Check the Limit Form

Apply L'Hopital's Rule

Differentiate the Terms

Simplify the Expression

Substitute $x = 1$

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Indeterminate Trap

The Power of L'Hopital's Rule

The Grand Simplification

The Final Reveal

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