Isa Trevino

2021-10-22

Calculate the following limits using the factorization formula
${x}^{n}-{a}^{n}=1x-a21{x}^{n-1}+a{x}^{n-2}+{a}^{2}{x}^{n-3}+\dots +{a}^{n-2}x+{a}^{n-1}2$, where n is a positive integer a is a real number.
$\underset{x\to a}{lim}\frac{{x}^{5}-{a}^{2}}{x-a}$

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Step 1
Given: $\underset{x\to a}{lim}\frac{{x}^{5}-{a}^{2}}{x-a}$
Step 2
Explanation:
$\underset{x\to a}{lim}\frac{{x}^{5}-{a}^{2}}{x-a}=\underset{x\to a}{lim}\frac{\left(x-a\right)\left({x}^{5-1}+a{x}^{5-2}+{a}^{2}{x}^{5-3}+{a}^{3}{x}^{5-4}+{a}^{4}{x}^{5-5}\right)}{\left(x-a\right)}$
$=\underset{x\to a}{lim}\left({x}^{4}+a{x}^{3}+{a}^{2}{x}^{2}+{a}^{3}x+{a}^{4}\right)$
$={a}^{4}+a{\left(a\right)}^{3}+{a}^{2}{\left(a\right)}^{2}+{a}^{3}\left(a\right)+{a}^{4}$
$={a}^{4}+{a}^{4}+{a}^{4}+{a}^{4}+{a}^{4}$
$=5{a}^{4}$
Hence, $\underset{x\to a}{lim}\frac{{x}^{5}-{a}^{5}}{x-a}=5{a}^{4}$
Step 3
Answer: $\underset{x\to a}{lim}\frac{{x}^{5}-{a}^{5}}{x-a}=5{a}^{4}$

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