The image contains mathematical problems requiring the application of L'Hôpital's Rule. Here's the step-by-step solution for the limits provided:

Problem 2:

$\lim_{x \to \infty} \frac{x + 3}{\sqrt{x^2 - 5} - 2}$

Solution:

1. Simplify the denominator: $\sqrt{x^2 - 5} \approx \sqrtThe{x image^ you2 uploaded} contains mathematical = x expressions \ involvingquad limits \ and L'Htext{(for large } x\text{)}.$

   Substituting this approximation:

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The first part of the image includes L'Hopital's Rule:

L'Hopital's Rule is used to evaluate limits of indeterminate forms like 0/0 or ∞/∞. The rule states: [ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ if the limit on the right side exists.

Now, let's analyze the problems:

1) (\lim_{x \to 3} \frac{x+3}{\sqrt{x^2 - }5} - 2}\ \approx \frac{x + 3}{x - 2}.

]

2.) We are looking for the limit as $x Divide$ approaches 3 for the expression $\frac{x+3}{\sqrt{x^2 - 5} - 2}$. Since direct substitution results numerator in a $\frac{0}{0}$ indeterminate form, L'Hopital and's Rule should be applied denominator.

2) $\lim_{x \to 0} \frac{5x^4 - 4x^2 - 1}{10 - x - 9x^3}$

For this limit, direct substitution also yields by $x$ to simplify: $\frac{x + 3}{x - 2} = \frac{1 + \frac{3}{x}}{1 - \frac{2}{x}}.$

3. As a \(\frac{0}{0} form, so applying L'Hopital'sx Rule is necessary.

\to \infty),Let me the terms now work through the \ steps of solving these limits. Would(\ youfrac like{ me to solve them in3 detail for you?}{x}) and $\frac{2}{x}$ approach 0: $\lim_{x \to \infty} \frac{1 + \frac{3}{x}}{1 - \frac{2}{x}} = \frac{1 + 0}{1 - 0} = 1.$

Final Answer: $\lim_{x \to \infty} \frac{x + 3}{\sqrt{x^2 - 5} - 2} = 1.$

Problem 4:

$\lim_{x \to \infty} \frac{5x^4 - 4x^2 - 1}{10 - x - 9x^3}$

Solution:

1. Identify the highest power of $x$ in the numerator and denominator:

   • Numerator: $5x^4$,
   • Denominator: $-9x^3$.

2. Factor out $x^4$ from the numerator and $x^3$ from the denominator: $\frac{5x^4 - 4x^2 - 1}{10 - x - 9x^3} = \frac{x^4\left(5 - \frac{4}{x^2} - \frac{1}{x^4}\right)}{x^3\left(-9 + \frac{10}{x^3} - \frac{1}{x^2}\right)}.$

3. Simplify by canceling $x^3$: $\frac{5x^4}{-9x^3} = \frac{5x}{-9} \text{ with the remaining terms approaching 0 as } x \to \infty.$

4. As $x \to \infty$:

   • The numerator is dominated by $5x^4$,
   • The denominator is dominated by $-9x^3$,
   • The other terms vanish.

Final Answer: $\lim_{x \to \infty} \frac{5x^4 - 4x^2 - 1}{10 - x - 9x^3} = \infty.$

Would you like more detailed steps or explanations for these problems?

Related Questions:

1. What is the application of L'Hôpital's Rule in indeterminate forms?
2. How does factoring help in simplifying limits as $x \to \infty$?
3. When do we approximate terms with the highest degree of $x$?
4. How can we use dominant terms to estimate limits more efficiently?
5. Can L'Hôpital's Rule be applied directly to problem 4?

Tip: Always analyze the dominant terms when solving limits as $x \to \infty$. This simplifies complex expressions significantly.