Is this limit right?

[sigma]

I want to see if I computed the following limit correctly. In a curve sketching question, I'm suppose to to find the horizontal asymptotes of \(\displaystyle \
\L\
f(x) = \frac{{2x^3 + x^2 + 1}}{{x^2 }}
\\)

So this is what I did.
\(\displaystyle \
\L\
\lim\limits_{x \to \pm \infty } \frac{{2x^3 + x^2 + 1}}{{x^2 }} = \lim\limits_{x \to \pm \infty } \frac{{\frac{{2x^3 }}{{x^2 }} + \frac{{x^2 }}{{x^2 }} + \frac{1}{{x^2 }}}}{{\frac{{x^2 }}{{x^2 }}}} = \lim\limits_{x \to \pm \infty } \frac{{2x + 1 + \frac{1}{{x^2 }}}}{1} = \frac{{0 + 1 + 0}}{1} = \frac{1}{1} = 1
\\)

Therefore, y = 1 is a horizontal asymptote.

Is that right?

 

[galactus]

You made an error. You have your limit approaching 0.

One can make an easy observation without taking the limit.

Since the degree of the numerator is greater than the degree of the

denominator, there is no horizontal asymptote.


Using long division on your function we get:

\(\displaystyle \L\\2x+\frac{1}{x^{2}}+1\)

It's easy to see this approaches infinity as x approaches infinity

 

[sigma]

[I am disappointed in my work]!