Question

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Answer 1

Answer:

-1/8

Step-by-step explanation:

lim x approaches -6     (sqrt( 10-x) -4) / (x+6)

Rationalize

   (sqrt( 10-x) -4)      (sqrt( 10-x) +4)

    ------------------- * -------------------

       (x+6)                 (sqrt( 10-x) +4)

We know ( a-b) (a+b) = a^2 -b^2

a= ( sqrt(10-x)   b = 4    

(10-x) -16

-------------------

(x+6) (sqrt( 10-x) +4)    

-6-x

-------------------

(x+6) (sqrt( 10-x) +4)

Factor out -1 from the numerator

-1( x+6)

-------------------

(x+6) (sqrt( 10-x) +4)

Cancel x+6 from the numerator and denominator

-1

-------------------

(sqrt( 10-x) +4)

Now take the limit

lim x approaches -6    -1/ (sqrt( 10-x) +4)

                                      -1/ (sqrt( 10- -6) +4)

                                      -1/ (sqrt(16) +4)

                                      -1 /( 4+4)

                                        -1/8

Answer 2

Answer:

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

Step-by-step explanation:

So first, we should always try direct substitution:

$\lim_{x \to -6}\frac{\sqrt{10-x}-4}{x+6}$

Plug -6 in for x:

$\frac{\sqrt{10-(-6)}-4}{(-6)+6} \\=\frac{\sqrt{16}-4}{-6+6}\\ =\frac{4-4}{-6+6}=0/0$

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

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. Recall the difference of two squares formula:

$(a-b)(a+b)=a^2-b^2$

The expression in the numerator is:

$\sqrt{10-x}-4$

Therefore, to cancel it out, we need to multiply by:

$\sqrt{10-x}+4$

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

$\lim_{x \to -6}\frac{\sqrt{10-x}-4}{x+6}\cdot\frac{\sqrt{10-x}+4}{\sqrt{10-x}+4}$

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

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

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

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

Simplify:

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

Factor out a negative 1 from the numerator:

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

The (x+6)s cancel out:

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

Now, plug -6 again:

$\frac{-1}{(\sqrt{10-(-6)}+4)}\\=\frac{-1}{\sqrt{16+4}}\\ =-1/(4+4)=-1/8=-.125$

Therefore:

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