To evaluate the given limit:

$\lim_{x \to 5^+} \left( \frac{1}{x-5} - \frac{1}{\sqrt{x-5}} \right),$

we note that both terms approach $\infty$ as $x \to 5^+$, leading to an indeterminate form of $\infty - \infty$. To resolve this, we will rewrite the expression with a common denominator and apply L'Hôpital's Rule if applicable.

Step 1: Rewrite with a common denominator

The expression can be rewritten as: $\frac{1}{x-5} - \frac{1}{\sqrt{x-5}} = \frac{\sqrt{x-5} - (x-5)}{(x-5)\sqrt{x-5}}.$

Step 2: Simplify the numerator

Simplify $\sqrt{x-5} - (x-5)$: $\sqrt{x-5} - (x-5) = \sqrt{x-5} - x + 5.$

Thus, the limit becomes: $\lim_{x \to 5^+} \frac{\sqrt{x-5} - x + 5}{(x-5)\sqrt{x-5}}.$

Step 3: Substitute $u = x - 5$

Let $u = x - 5$, so as $x \to 5^+$, $u \to 0^+$. The expression now becomes: $\lim_{u \to 0^+} \frac{\sqrt{u} - u}{u\sqrt{u}} = \lim_{u \to 0^+} \frac{1 - \sqrt{u}}{u}.$

Step 4: Apply L'Hôpital's Rule

The form $\frac{1 - \sqrt{u}}{u}$ is indeterminate ($\frac{0}{0}$) as $u \to 0^+$, so we apply L'Hôpital's Rule by differentiating the numerator and denominator:

• Derivative of $1 - \sqrt{u}$ is $-\frac{1}{2\sqrt{u}}$,
• Derivative of $u$ is $1$.

Thus, the limit becomes: $\lim_{u \to 0^+} -\frac{\frac{1}{2\sqrt{u}}}{1} = \lim_{u \to 0^+} -\frac{1}{2\sqrt{u}}.$

Step 5: Evaluate the limit

As $u \to 0^+$, $\sqrt{u} \to 0^+$, so $-\frac{1}{2\sqrt{u}} \to -\infty$.

Final Answer:

$\lim_{x \to 5^+} \left( \frac{1}{x-5} - \frac{1}{\sqrt{x-5}} \right) = -\infty.$

Would you like further clarification or detailed steps? Here are related questions to explore:

1. What is L'Hôpital's Rule and when is it applicable?
2. How do you rewrite indeterminate forms involving subtraction?
3. How can substitution simplify limits with square roots?
4. Why does $-\infty$ appear in this limit?
5. How can we verify the correctness of this approach?

Tip: Always simplify the expression before applying L'Hôpital's Rule to avoid unnecessary complexity!