Question

evaluate the limit:

Answer 1

\[\lim_{x \rightarrow \infty} \frac{ x ^{2} + 3x -2 }{3x ^{2} +4x -1 }\]

Answer 2

Looks like a case where L'Hopital's rule will be useful.

Always try to evaluate it straight up first. If we try that, we'll just get
\[ \frac{\infty}{\infty} \]

So let's try L'Hopital. Take the derivative of the numerator and denomerator, independently. This will give us
\[\lim_{x->\infty} \frac{2x+3}{6x+4}\]

Onece again, try to evaluate. We'll still get
\[ \frac{\infty}{\infty} \]

Use L'Hopital again:
\[\lim_{x->\infty} \frac{2}{6}\]

Alas! Evaluate the limit now to get
\[\lim_{x->\infty} \frac{2}{6} = \frac{2}{6} = \frac{1}{3} \]

Answer 3

A perhaps more clever way of doing this is to divide the entire limit piece by x^2, so we'd have

\[ \lim_{ x-> \infty} \frac{1 + \frac{3}{x} - \frac{2}{x^2}}{3 + \frac{4}{x} -\frac{1}{x^2}} \]

Since
\[ \lim_{ x-> \infty} \frac{\text{integer}}{x} = 0\] and
\[ \lim_{ x-> \infty} \frac{\text{integer}}{x^2} = 0\]

We easily see that the limit is 1/3.

Answer 4

yeah i see thanks man .. u help alot .. thanks