Starting with limits: lim [x -> infty] [(5x + 2)/(x - 1)]

[jwpaine]

Find the limit of \(\displaystyle \L \frac{5x+2}{x-1}\) when x approaches \(\displaystyle \infty\)

divide all terms in the numerator and denominator by x

\(\displaystyle \L \frac{5+\frac{2}{x}}{1-\frac{1}{x}}\)

The limit of \(\displaystyle \L \frac{2}{x} and \frac{1}{x}\) is 0..so the limit of

\(\displaystyle \L \frac{5x+2}{x-1} = \frac{5+\frac{2}{x}}{1-\frac{1}{x}} = \frac{5+0}{1-0} = 5\)

Have I done this correctly?

Thanks,
John.

 

[galactus]

Yes, JW, you're correct.

Another thing you could do is divide and get:

\(\displaystyle \L\\\frac{7}{x-1}+5\)

Now, \(\displaystyle \L\\\lim_{x\to\infty}\frac{7}{x-1}=0\)

Therefore, 0+5=5.

Just another way.

Besides, a good thing to know when it comes to limits of rational expressions is we can eliminate all but the largest powers of x.

Doing this, we have \(\displaystyle \L\\\frac{5x}{x}=5\)