The first thing one would

**try** is to just put x= 9/8. Of course, that doesn't work here (it never does for interesting problems) but it shows where the problem is- the denominator becomes 0 while the numerator does not. Now, what happens if x is

**very close** to 9/8 but just slightly less? You can answer that by actually putting in such a number. 9/8= 1.125 so what happens if x= 1.120? In that case, the numerator is -22x= -24.26 and the denominator is 9- 8(1.120)= 9- 8.96= .04. -24.26/.04= -606.5. In other words, the fraction itself is a very large

**negative** number- since the denominator is going to 0 and the numerator isn't, the limit will be "negative infinity". You could also do that by recognizing that $\displaystyle 9- 8x= -8\left(x- \frac{9}{8}\right)$. If x is

**less** than 9/8, the quantity in the parentheses, $\displaystyle x- \frac{9}{8}$ is negative so $\displaystyle -8\left(x- \frac{9}{8}\right)$ is the product of two negative numbers and so is negative. Since the numerator, -22x, is also negative, the fraction is negative and the limit is negative infinity. For x

**greater** than 9/8, $\displaystyle x- \frac{9}{8}$ is

**positive** so that the denominator is negative and the fraction is the quotient of

**two** negative numbers. The fraction is positive so the limit is

**positive** infinity.

For $\displaystyle \sqrt{x^2+ 4x+ 1}- x$ multiply by the fraction $\displaystyle \frac{\sqrt{x^2+ 4x+ 1}+ x}{\sqrt{x^2+ 4x+ 1}+ x}$ then do thesame thing.

The horizontal asymptotes of the curve are, of course, the values of y as x goes to plus or minus infinity. I recommend that you divide both numerator and denominator by x. That way the numerator is the constant 18. What happens in the denominator?