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Evaluate the limit if possible. Walk through the process

To solve the limit 

\[
\lim_{x \to 0} \frac{\sqrt{x + 5} - \sqrt{5}}{x},
\]

we need to evaluate it carefully, as directly substituting \( x = 0 \) would result in an indeterminate form of \(\frac{0}{0}\). Here's a step-by-step approach:

### Step 1: Multiply by the Conjugate

To simplify the expression, multiply the numerator and denominator by the conjugate of the numerator:

\[
\frac{\sqrt{x + 5} - \sqrt{5}}{x} \cdot \frac{\sqrt{x + 5} + \sqrt{5}}{\sqrt{x + 5} + \sqrt{5}}.
\]

This simplifies to:

\[
\frac{(\sqrt{x + 5} - \sqrt{5})(\sqrt{x + 5} + \sqrt{5})}{x(\sqrt{x + 5} + \sqrt{5})}.
\]

### Step 2: Simplify the Numerator

Using the difference of squares in the numerator:

\[
(\sqrt{x + 5} - \sqrt{5})(\sqrt{x + 5} + \sqrt{5}) = (x + 5) - 5 = x.
\]

Now, the expression becomes:

\[
\frac{x}{x(\sqrt{x + 5} + \sqrt{5})} = \frac{1}{\sqrt{x + 5} + \sqrt{5}}.
\]

### Step 3: Evaluate the Limit

Now, substitute \( x = 0 \):

\[
\lim_{x \to 0} \frac{1}{\sqrt{x + 5} + \sqrt{5}} = \frac{1}{\sqrt{5} + \sqrt{5}} = \frac{1}{2\sqrt{5}}.
\]

### Final Answer

\[
\lim_{x \to 0} \frac{\sqrt{x + 5} - \sqrt{5}}{x} = \frac{1}{2\sqrt{5}}.
\]

Would you like further details on any step?

### Related Questions:
1. How do you handle limits involving square roots and indeterminate forms?
2. What is the difference between an indeterminate form and a defined limit?
3. How does multiplying by the conjugate help simplify expressions with radicals?
4. Can L'Hopital's rule be used as an alternative method for this limit?
5. How can you verify the correctness of a limit evaluation?

### Tip:
Multiplying by the conjugate is a powerful technique when dealing with limits involving square roots to eliminate indeterminate forms.

Evaluate the limit if possible: lim (x -> 0) (sqrt(x + 5) - sqrt(5)) / x.

Learn how to solve the limit lim (x -> 0) (sqrt(x + 5) - sqrt(5)) / x using the conjugate method. This step-by-step solution walks through simplifying the expression and handling indeterminate forms, suitable for Pre-Calculus or Calculus students.

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