The limit problem in the image is:

$\lim_{{x \to 2}} \frac{x^2}{x^4 - 16}$

Let's solve this step-by-step:

Step 1: Direct Substitution

First, substitute $x = 2$ into the expression:

$\frac{2^2}{2^4 - 16} = \frac{4}{16 - 16} = \frac{4}{0}$

This results in a division by zero, which means the direct substitution does not work. We need to simplify the expression further.

Step 2: Factor the Denominator

The denominator is $x^4 - 16$, which is a difference of squares:

$x^4 - 16 = (x^2 - 4)(x^2 + 4)$

We can further factor $x^2 - 4$ as:

$x^2 - 4 = (x - 2)(x + 2)$

So the denominator becomes:

$x^4 - 16 = (x - 2)(x + 2)(x^2 + 4)$

Step 3: Simplify the Expression

The limit expression now becomes:

$\lim_{{x \to 2}} \frac{x^2}{(x - 2)(x + 2)(x^2 + 4)}$

Since $x - 2$ is a factor in the denominator, we cannot directly substitute $x = 2$. But we can attempt to simplify it further.

Step 4: Use L'Hopital's Rule

Since direct substitution leads to an indeterminate form of $\frac{0}{0}$, we can apply L'Hopital's Rule, which states that if $\lim_{{x \to a}} \frac{f(x)}{g(x)}$ results in $\frac{0}{0}$, we can differentiate the numerator and denominator and then take the limit.

• Differentiate the numerator: $\frac{d}{dx}(x^2) = 2x$
• Differentiate the denominator: $\frac{d}{dx}((x - 2)(x + 2)(x^2 + 4))$

I'll calculate that.After applying L'Hopital's Rule and differentiating both the numerator and the denominator, the limit as $x \to 2$ is:

$\lim_{{x \to 2}} \frac{x^2}{x^4 - 16} = \frac{1}{8}$

Would you like more details on the differentiation steps, or do you have any other questions?

Here are some related questions you might find helpful:

1. How does L'Hopital's Rule work, and when can you apply it?
2. What are indeterminate forms, and why do they arise in limits?
3. Can we solve this limit without using L'Hopital's Rule?
4. How do you factor higher-order polynomials like $x^4 - 16$?
5. What is the significance of the result $\frac{1}{8}$ in this context?

Tip: Always check if a limit results in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hopital's Rule.