Question

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

$\huge{\underline{\underline{\mathcal\color{fuchsia}{\bigstar{Answer}}}}}$

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

Step-by-step explanation:

$\mapsto{\sf{   }}$:So first, we should always try direct substitution:

$\bold{↬{ }}$:$\begin{gathered}\lim_{x \to -6}\dfrac{\sqrt{10-x}-4}{x+6} \\\end{gathered}$

$\mapsto{\sf{   }}$:Plug -6 in for x:

$\begin{gathered}\dfrac{\sqrt{10-(-6)}-4}{(-6)+6} \\=\dfrac{\sqrt{16}-4}{-6+6}\\ =\dfrac{4-4}{-6+6}=0/0\end{gathered}$

$\rightsquigarrow$:This is the indeterminate form. This doesn't mean the limit does not exist, but rather we need to simplify it first.

$\rightsquigarrow$:Looking at the limit, we see that there is a square root in the numerator. Therefore, we can use the difference of two squares to cancel out the square root in the numerator.

$\rightsquigarrow$:Recall the difference of two squares formula:

$\mapsto{\sf{   }}$:$\green{(a-b)(a+b)=a^2-b^2}$

$\bold{↬{ }}$:The expression in the numerator is:

$\sqrt{10-x}-4$

$\mapsto{\sf{   }}$:Therefore, to cancel it out, we need to multiply by:

$\sqrt{10-x}+4$

$\rightsquigarrow$ Essentially, you just change the sign. So, multiply both the numerator and denominator by this expression:

$\begin{gathered}\lim_{x \to -6}\dfrac{\sqrt{10-x}-4}{x+6}\cdot\dfrac{\sqrt{10-x}+4}{\sqrt{10-x}+4} \\\end{gathered}$

For the numerator, this is the difference of two squares pattern. Therefore:

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

The roots in the numerator cancel. 4 squared is 16.

Simplify:

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

Simplify:

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

Factor out a negative 1 from the numerator:

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

The (x+6)s cancel out:

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

Now, plug -6 again:

$\begin{gathered}\dfrac{-1}{(\sqrt{10-(-6)}+4)}\\=\dfrac{-1}{\sqrt{16+4}}\\ =\dfrac{-1}{(4+4)}=\dfrac{-1}{8}=-.125\end{gathered}$

Therefore:

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

Answer 2

EXPLANATION.

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

As we know that,

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

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

$\sf \implies \lim_{x \to - 6} \dfrac{4 - 4}{- 6 + 6}.$

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

As we can see that it is in the form of 0/0.

We can simply factorizes the equation.

But if root is in 0/0 form we can simply rationalizes the equation, we get.

Rationalizes the equation, we get.

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

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

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

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

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

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

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

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

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

$\sf \implies \lim_{x \to - 6} \dfrac{- 1}{8}.$

$\sf \implies values \ of \ equation \lim_{x \to - 6} \dfrac{\sqrt{10 - x} - 4}{x + 6} = \dfrac{- 1}{8}.$