Question

✓ 4x + 36 - 6
Evaluate the limit: lim
20

Answer 1

$\qquad$ ☀️$\pink{\bf{ {Answer = \: \: \dfrac{1}{3} }}}$

$\qquad$$\pink{\bf\twoheadrightarrow \bf\displaystyle \lim_{x \to 0} \dfrac{\sqrt{4x +36} -6}{x }}$

$\bf\twoheadrightarrow \displaystyle \lim_{x \to 0} \bigg(\dfrac{\sqrt{4x+36} - 6}{x} \bigg) \ \times \bigg( \dfrac{\sqrt{4x+36} +6}{\sqrt{4x+36} +6} \bigg)$

$\bf \twoheadrightarrow \displaystyle \lim_{x \to 0} \bigg(\dfrac{(\sqrt{4x+36} - 6)\times (\sqrt{4x+36}+6)}{x(\sqrt{4x+36}+6)} \bigg)$

$\bf \twoheadrightarrow \displaystyle \lim_{x \to 0} \bigg(\dfrac{(\sqrt{4x+36}) ^{2} - (6)^{2} }{x(\sqrt{4x+36} +6)} \bigg)$

$\bf \twoheadrightarrow \displaystyle \lim_{x \to 0} \bigg(\dfrac{4x+36-36}{x(\sqrt{4x+36} +6)} \bigg)$

$\bf \twoheadrightarrow \displaystyle \lim_{x \to 0} \bigg(\dfrac{4x+\cancel{36}-\cancel{36}}{x(\sqrt{4x+36} +6)} \bigg)$

$\bf \twoheadrightarrow \displaystyle \lim_{x \to 0} \bigg(\dfrac{4\cancel{x}}{\cancel{x}(\sqrt{4x+36} +6)} \bigg)$

$\bf \twoheadrightarrow \displaystyle \lim_{x \to 0} \bigg(\dfrac{4}{(\sqrt{4x+36} +6)} \bigg)$

Now, substitute x = 0

$\qquad$ $\bf \twoheadrightarrow \bigg(\dfrac{4}{(\sqrt{4\times 0+36} +6)} \bigg)$

$\qquad$ $\sf \twoheadrightarrow \bigg(\dfrac{4}{(\sqrt{36}+6)} \bigg)$

$\qquad$ $\sf\twoheadrightarrow \bigg(\dfrac{4}{(6+6)} \bigg)$

$\qquad$ $\sf \twoheadrightarrow \bigg(\dfrac{4}{12} \bigg)$

$\qquad$ $\pink{\bf \twoheadrightarrow \dfrac{1}{3} }$

Answer 2

Rationalizing the function, it is found that the limit is $\frac{1}{3}$.

What is the limit?

The limit that we want to find is:

$\lim_{x \rightarrow 0} \frac{\sqrt{4x + 36} - 6}{x}$

Using the standard procedure of replacing x by 0, we end with a division by 0, hence it cannot be used.

When there is a division by 0 and a square root, we can rationalize the function, as follows:

$\frac{\sqrt{4x + 36} - 6}{x} \times \frac{\sqrt{4x + 36} + 6}{\sqrt{4x + 36} + 6} = \frac{4x + 36 - 36}{x(\sqrt{4x + 36} + 6)} = \frac{4}{\sqrt{4x + 36} + 6}$

Then:

$\lim_{x \rightarrow 0} \frac{\sqrt{4x + 36} - 6}{x} = \lim_{x \rightarrow 0} \frac{4}{\sqrt{4x + 36} + 6}$

Now the standard procedure of replacing can be applied, hence:

$\lim_{x \rightarrow 0} \frac{4}{\sqrt{4x + 36} + 6} = \frac{4}{12} = \frac{1}{3}$

The limit is $\frac{1}{3}$.

You can learn more about limits at https://brainly.com/question/26367086