Question

assignment 3: problem 25 (1 point) use factoring to calculate this limit $lim_{x \to a}\frac{x^{3}-a^{3}}{x^{5}-a^{5}}$ if you want a hint, try doing this numerically for a couple of values of x and a. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor

Explanation:

Step1: Use the formula $x^n - a^n=(x - a)(x^{n - 1}+x^{n - 2}a+\cdots+xa^{n - 2}+a^{n - 1})$

$x^{3}-a^{3}=(x - a)(x^{2}+xa + a^{2})$ and $x^{5}-a^{5}=(x - a)(x^{4}+x^{3}a+x^{2}a^{2}+xa^{3}+a^{4})$

Step2: Substitute the factored - forms into the limit

$\lim_{x
ightarrow a}\frac{x^{3}-a^{3}}{x^{5}-a^{5}}=\lim_{x
ightarrow a}\frac{(x - a)(x^{2}+xa + a^{2})}{(x - a)(x^{4}+x^{3}a+x^{2}a^{2}+xa^{3}+a^{4})}$

Step3: Cancel out the common factor $(x - a)$

Since $x
eq a$ when taking the limit, we can cancel out $(x - a)$ to get $\lim_{x
ightarrow a}\frac{x^{2}+xa + a^{2}}{x^{4}+x^{3}a+x^{2}a^{2}+xa^{3}+a^{4}}$

Step4: Substitute $x = a$ into the simplified expression

$\frac{a^{2}+a\cdot a + a^{2}}{a^{4}+a^{3}\cdot a+a^{2}\cdot a^{2}+a\cdot a^{3}+a^{4}}=\frac{3a^{2}}{5a^{4}}$

Step5: Simplify the result

$\frac{3a^{2}}{5a^{4}}=\frac{3}{5a^{2}}$ (assuming $a
eq0$)

Answer:

$\frac{3}{5a^{2}}$