Question

Evaluate limx→∞ √x^4-2x^3-6/3x^2+6

Answer 1

Is that ALL under a root sign?

Answer 2

Divide every term by the highest power in the denominator. The only thing you need to worry about when dividing out the terms in the square root is you need to rewrite the power of x you're dividing by as a square root. So the highest power of x in the denominator is x^2, so we divide every term by x^2. Now in order to divide the terms under the square root, we turn x^2 into sqrt(x^4), because these are essentially equivalent expressions. SO doing that gives us this:

\[\frac{ \sqrt{\frac{ x^{4} }{ x^{4} }-\frac{ 2x^{3} }{ x^{4} }-\frac{ 6 }{ x^{4} }} }{ \frac{ 3x ^{2} }{ x^{2} } +\frac{ 6 }{ x^{4} }}\]
assuming im reading it right o.o

Answer 3

I am still left with √-2x/3. What do I do next?

Answer 4

Did I write the problem correctly?

Answer 5

You wrote it correctly. I am still left with(√-2x) /3. The answer is not 0

Answer 6

Well, let's look at everything that cancels out by what I did above:



Now when you have this situation:

\[\lim_{x \rightarrow \infty}\frac{ x ^{a} }{ x^{b} }\] and b >a, the limit is 0. If a > b, the limit is infinity. So when we look at our problem, all of the fractions that are left behind all have the power of x greater in the denominator, so every singleone of those fractions becomes 0. Then your answer is just left behind afterwards.