Question

Evaluate the limit: $\pink{\sf \lim_{x \to -6} \dfrac{\sqrt{10 - x}-4 }{x + 6}}$ 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,

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

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

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

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

As we can see that,

It is in the form of 0/0 indeterminant.

If root exists in 0/0 form then rationalizes the equation, we get.

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

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

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

$\sf \implies \displaystyle \lim_{x \to -6} \bigg[\dfrac{10 - x - 16}{[x + 6][\sqrt{10 - x} + 4]} \bigg]$

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

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

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

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

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

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

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

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

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