aortiH

2021-11-01

Find the following limits or state that they do not exist. Assume a, b, c, and k are fixed real numbers.
$\underset{x\to 3}{lim}\frac{\frac{1}{{x}^{2}+2x}-\frac{1}{15}}{x-3}$

pivonie8

Concept:
The calculus helps in understanding the changes between values that are related by a function. It is used in different areas such as physics, biology and sociology and so on. The calculus is the language of engineers, scientists and economics.
Given:
$\underset{x\to 3}{lim}\frac{\frac{1}{{x}^{2}+2x}-\frac{1}{15}}{x-3}$
Solution:
Thus the given function is as follows,
$\underset{x\to 3}{lim}\frac{\frac{1}{{x}^{2}+2x}-\frac{1}{15}}{x-3}$
Computing the limit by simplifying the function as follows, $\underset{x\to a}{lim}\frac{{x}^{2}-{a}^{2}}{\sqrt{x}-\sqrt{a}}=\underset{x\to a}{lim}\left(\left(\sqrt{x}+\sqrt{a}\right)\left(x+a\right)\right)$
$\underset{x\to a}{lim}\frac{{x}^{2}-{a}^{2}}{\sqrt{x}-\sqrt{a}}=\underset{x\to a}{lim}\frac{\left(x-a\right)\left(x+a\right)}{\sqrt{x}-\sqrt{a}}$
$=\underset{x\to a}{lim}\frac{\left({\left(\sqrt{x}\right)}^{2}-{\left(\sqrt{a}\right)}^{2}\right)\left(x+a\right)}{\sqrt{x}-\sqrt{a}}$
$\underset{x\to a}{lim}\left(\left(\sqrt{x}+\sqrt{x}\right)\left(x+a\right)\right)$
$\underset{x\to a}{lim}\frac{{x}^{2}-{a}^{2}}{\sqrt{x}-\sqrt{a}}=\underset{x\to a}{lim}\left(\left(\sqrt{x}+\sqrt{a}\right)\left(x+a\right)\right)$
$=\left(\sqrt{a}+\sqrt{a}\right)\left(a+a\right)=2\sqrt{a}\left(2a\right)$
$=2a\sqrt{a}$

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