Question

My question was on limits and I didn't know how to post it in the correct format so I took a picture of it.
http://imgur.com/rvi2xLQ

Answer 1

does x=1 cause a divide by zero problem?

Answer 2

also, when x get really big .... everything but the largest exponents becomes minimal in effect

so it equates to x^2/sqrt(9x^6)

Answer 3

i think you have to do something to the equation

Answer 4

you can if you want, but there are a few different ways to approach limits, the common sense approach is good enough for these.

Answer 5

the first one, x=1 has no issues in the math, no divide by 0, and not negative sqrt so its just the value at x=1

Answer 6

the other can be assessed by dividing all the terms by sqrt(x^6) if need be

Answer 7

can you please explain the second limit?

Answer 8

\[\frac{x^3+1}{\sqrt{9x^6+x^2-1}}\]
lets multiply by 1/sqrt(x^6) / 1/sqrt(x^6), a useful form of 1
\[\frac{1/\sqrt{x^6}}{1/\sqrt{x^6}}\frac{x^3+1}{\sqrt{9x^6+x^2-1}}\]
\[\frac{\frac{x^3}{x^3}+\frac 1{x^3}}{\sqrt{\frac{9x^6}{x^6}+\frac{x^2}{x^6}-\frac{1}{x^6}}}\]
reduce
\[\frac{1+x^{-3}}{\sqrt{9+x^{-4}-x^{-6}}}\]
now, anything with a negative exponent, goes to zero, and we are left with?