To evaluate the limit:

$\lim_{x \to 1}\frac{\sqrt{17-x}-4}{\sqrt{10-x}-3}$

Step 1: Direct Substitution

Substituting $x = 1$:

$\sqrt{17 - 1} - 4 = \sqrt{16} - 4 = 4 - 4 = 0$

$\sqrt{10 - 1} - 3 = \sqrt{9} - 3 = 3 - 3 = 0$

Since we get an indeterminate form $\frac{0}{0}$, we proceed by rationalizing.

Step 2: Rationalization

We rationalize both the numerator and denominator.

Numerator:

$\sqrt{17-x} - 4$

Multiply and divide by its conjugate $\sqrt{17-x} + 4$:

$(\sqrt{17-x} - 4) \cdot \frac{\sqrt{17-x} + 4}{\sqrt{17-x} + 4}$

Using the identity $(a - b)(a + b) = a^2 - b^2$:

$\frac{(17-x) - 16}{\sqrt{17-x} + 4} = \frac{1-x}{\sqrt{17-x} + 4}$

Denominator:

$\sqrt{10-x} - 3$

Multiply and divide by its conjugate $\sqrt{10-x} + 3$:

$(\sqrt{10-x} - 3) \cdot \frac{\sqrt{10-x} + 3}{\sqrt{10-x} + 3}$

Using the identity $(a - b)(a + b) = a^2 - b^2$:

$\frac{(10-x) - 9}{\sqrt{10-x} + 3} = \frac{1-x}{\sqrt{10-x} + 3}$

Step 3: Simplification

Now, rewrite the fraction:

$\frac{\frac{1-x}{\sqrt{17-x} + 4}}{\frac{1-x}{\sqrt{10-x} + 3}}$

Cancel $1-x$:

$\frac{1}{\sqrt{17-x} + 4} \div \frac{1}{\sqrt{10-x} + 3} = \frac{\sqrt{10-x} + 3}{\sqrt{17-x} + 4}$

Step 4: Evaluate at $x = 1$

$\frac{\sqrt{10-1} + 3}{\sqrt{17-1} + 4} = \frac{\sqrt{9} + 3}{\sqrt{16} + 4} = \frac{3+3}{4+4} = \frac{6}{8} = \frac{3}{4}$

Final Answer:

$\frac{3}{4}$

Would you like a more detailed explanation of any step?

Related Questions:

1. How do you determine when to rationalize an expression in a limit problem?
2. What are alternative methods to evaluate limits when direct substitution gives $\frac{0}{0}$?
3. How does L'Hôpital's Rule apply to similar limit problems?
4. Can you generalize this technique for similar square root limits?
5. How does this problem relate to derivatives?

Tip:

When dealing with limits involving square roots, rationalization is often the best first step to simplify indeterminate forms.