Use analytical methods to evaluate each of the following limits. If a limit does not
exist, explain why. Assign ∞ or −∞ to the limit where appropriate.
lim (√[2 − x] − 1) / [3 − √(4x + 5)]
x→1

We are tasked with evaluating the limit:

\[
\lim_{x \to 1} \frac{\sqrt{2 - x} - 1}{3 - \sqrt{4x + 5}}
\]

### Step 1: Substitution of \( x = 1 \)

First, let's substitute \( x = 1 \) into the expression to see if we get an indeterminate form or a specific value:

\[
\frac{\sqrt{2 - 1} - 1}{3 - \sqrt{4(1) + 5}} = \frac{\sqrt{1} - 1}{3 - \sqrt{9}} = \frac{1 - 1}{3 - 3} = \frac{0}{0}
\]

Since this results in the indeterminate form \( \frac{0}{0} \), we need to apply **L'Hopital's Rule** or other analytical methods.

### Step 2: Apply L'Hopital's Rule

L'Hopital's Rule states that if the limit results in \( \frac{0}{0} \), we can differentiate the numerator and the denominator separately and then take the limit again. Let's proceed by differentiating the numerator and denominator.

#### Numerator: \( f(x) = \sqrt{2 - x} - 1 \)

The derivative of the numerator \( f(x) = \sqrt{2 - x} - 1 \) is:

\[
f'(x) = \frac{d}{dx} \left( \sqrt{2 - x} \right) = \frac{-1}{2\sqrt{2 - x}}
\]

#### Denominator: \( g(x) = 3 - \sqrt{4x + 5} \)

The derivative of the denominator \( g(x) = 3 - \sqrt{4x + 5} \) is:

\[
g'(x) = \frac{d}{dx} \left( 3 - \sqrt{4x + 5} \right) = - \frac{4}{2\sqrt{4x + 5}} = \frac{-2}{\sqrt{4x + 5}}


Use analytical methods to evaluate each of the following limits. If a limit does not exist, explain why. Assign ∞ or −∞ to the limit where appropriate. lim (√[2 − x] − 1) / [3 − √(4x + 5)], x→1

Learn how to evaluate the limit lim (sqrt(2 - x) - 1) / (3 - sqrt(4x + 5)) as x approaches 1 using L'Hopital's Rule. This solution involves differentiation of radical functions and the application of L'Hopital's Rule. Suitable for undergraduate calculus or advanced high school students.

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