Question

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Evaluate the limit:
$\lim_{x \to -6} \frac{\sqrt{10 - x}-4 }{x + 6}$
Please explain how to rationalize and solve for the limit.​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{\sqrt{10 - x} - 4}{x + 6}$

As we know that,

First we put the value of x = -6 in equation and check their indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{\sqrt{10 - (-6)} - 4}{(-6) + 6}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{\sqrt{16} - 4}{0}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{4 - 4}{0} = \dfrac{0}{0}$

As we can see that it is a form of 0/0 indeterminant form.

in 0/0 form if root is exist in equation, simply rationalize the equation, we get.

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{\sqrt{10 - x} - 4}{x + 6} \ \times \ \dfrac{\sqrt{10 - x} + 4}{\sqrt{10 - x}+ 4 }$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{(\sqrt{10 - x} - 4)(\sqrt{10 - x} + 4)}{(x + 6)(\sqrt{10 - x} + 4)}$

As we know that,

Formula of :

⇒ x² - y² = (x - y)(x + y).

Using this formula in equation, we get.

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{(\sqrt{10 - x)^{2} } - (4)^{2} }{(x + 6)(\sqrt{10 - x} + 4)}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{10 - x - 16}{(x + 6)(\sqrt{10 - x} + 4)}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{- x - 6}{(x + 6)(\sqrt{10 - x} + 4)}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{-(x + 6)}{(x + 6)(\sqrt{10 - x} + 4)}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{- 1}{(\sqrt{10 - x} + 4)}$

Put the value of x = -6 in equation, we get.

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{-1}{(\sqrt{10 - (-6)} + 4)}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{-1}{\sqrt{10 + 6} + 4}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{-1}{\sqrt{16} + 4}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{-1}{4 + 4} = \dfrac{-1}{8}$

$\sf \implies \displaystyle \lim_{x \to -6} \dfrac{\sqrt{10 - x} - 4}{x + 6} = \dfrac{-1}{8}$

                                                                                                                         

MORE INFORMATION.

By using some standard expansion.

(1) = eˣ = 1 + x + x²/2! + x³/3! + ,,,,,,,

(2) = e⁻ˣ = 1 - x + x²/2! - x³/3! + ,,,,,,,,

(3) = ㏒(1 + x) = x - x²/2 + x³/3 - ,,,,,,,,,

(4) = ㏒(1 - x) = -x - x²/2 - x³/3 - ,,,,,,,,

(5) = aˣ = 1 + (x㏒ a) + (x ㏒ a)²/2! + (x ㏒ a)³/3! + ,,,,,,

Answer 2

REFER to the attachment

hopefully it will work ☺️☺️