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

Skilled2021-11-02Added 91 answers

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,$\underset{x\to a}{lim}\frac{{x}^{2}-{a}^{2}}{\sqrt{x}-\sqrt{a}}=\underset{x\to a}{lim}\left((\sqrt{x}+\sqrt{a})(x+a)\right)$

$\underset{x\to a}{lim}\frac{{x}^{2}-{a}^{2}}{\sqrt{x}-\sqrt{a}}=\underset{x\to a}{lim}\frac{(x-a)(x+a)}{\sqrt{x}-\sqrt{a}}$

$=\underset{x\to a}{lim}\frac{({\left(\sqrt{x}\right)}^{2}-{\left(\sqrt{a}\right)}^{2})(x+a)}{\sqrt{x}-\sqrt{a}}$

$\underset{x\to a}{lim}\left((\sqrt{x}+\sqrt{x})(x+a)\right)$

$\underset{x\to a}{lim}\frac{{x}^{2}-{a}^{2}}{\sqrt{x}-\sqrt{a}}=\underset{x\to a}{lim}\left((\sqrt{x}+\sqrt{a})(x+a)\right)$

$=(\sqrt{a}+\sqrt{a})(a+a)=2\sqrt{a}\left(2a\right)$

$=2a\sqrt{a}$

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:

Solution:

Thus the given function is as follows,

Computing the limit by simplifying the function as follows,