Question

if ( a
eq 0 ), then ( limlimits_{x \to a} \frac{x^2 - a^2}{x^4 - a^4} ) is

Explanation:

Step1: Factor numerator and denominator

The numerator \(x^2 - a^2\) can be factored as \((x - a)(x + a)\) using the difference of squares formula \(A^2 - B^2=(A - B)(A + B)\). The denominator \(x^4 - a^4\) is also a difference of squares, where \(x^4=(x^2)^2\) and \(a^4=(a^2)^2\), so \(x^4 - a^4=(x^2 - a^2)(x^2 + a^2)\). Then we can further factor \(x^2 - a^2\) in the denominator as \((x - a)(x + a)\) as well. So the function becomes:
\[
\frac{(x - a)(x + a)}{(x - a)(x + a)(x^2 + a^2)}
\]

Step2: Cancel common factors

Since \(x\to a\) and \(a
eq0\), when \(x
eq a\) (because we are taking the limit as \(x\) approaches \(a\), not evaluating at \(x = a\)), we can cancel out the common factors \((x - a)\) and \((x + a)\) from the numerator and the denominator. After canceling, we are left with:
\[
\frac{1}{x^2 + a^2}
\]

Step3: Evaluate the limit

Now we can find the limit as \(x\) approaches \(a\) by substituting \(x = a\) into the simplified function \(\frac{1}{x^2 + a^2}\). So we have:
\[
\lim_{x\to a}\frac{1}{x^2 + a^2}=\frac{1}{a^2 + a^2}=\frac{1}{2a^2}
\]

Answer:

\(\frac{1}{2a^2}\)