Question

Evaluate the limit using the appropriate Limit Law(s). (If an answer does not exist, enter DNE.) lim x→8 1 + 3 x 5 − 6x2 + x3

Answer 1

Answer:

$[tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)$=$1-384 +512+3\sqrt{8} -18(8)^{5/2} +3 (8)^{7/2} =1223.601$

And the limit on this case exists.

Step-by-step explanation:

We want to find the following limit:

$lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)$

First we can distribute the polynomials like this:

$lim_{x \to 8} (1-6x^2 +x^3+3\sqrt{x} -18 x^{5/2} +3x^{7/2})$

And Now we can use the distributive property for the limit and we got:

$lim_{x \to 8} 1 - 6 lim_{x \to 8} x^2 + lim_{x \to 8} x^3 +3 lim_{x \to 8} \sqrt{x} -18 lim_{x \to 8} x^{5/2} + 3 lim_{x \to 8} x^{7/2}$

And now we can evaluate the limit and we got:

$[tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)$=$1-384 +512+3\sqrt{8} -18(8)^{5/2} +3 (8)^{7/2} =1223.601$

And the limit on this case exists.