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 Question from Jordan, a student: OK, I am having trouble figuring out how to solve this. lim as x approaches infinite of 5x + 2/x-1 Do i need to take a sin or cos or what? If can send me a step by step that would be so informative.

We have two responses for you

Jordon,

Why not split it up?

lim a+b+c = lim a + lim b + lim c

So lim 5x + 2/x - 1 = lim 5x + lim 2/x - lim 1.

Perhaps you forgot parentheses? If so, where should they be?

Stephen.

Hi Jordan,

The fact you need to use here is that

Here is a similar problem.

lim as x approaches infinite of (3x - 8)/(4 - 7x)

The technique is to divide the numerator and denominator by x. The numerator becomes

(3x - 8)/x = 3x/x - 8/x = 3 - 8 × 1/x

and the denominator becomes

(4 - 7x)/x = 4 × 1/x - 7.

As x approaches infinity 1/x approaches zero so the numerator approaches 3 - 0 = 3 and the denominator approaches 0 - 7 = -7. Thus as x approaches infinity the fraction (3x - 8)/(4 - 7x) approaches -3/7.

I hope this helps,
Harley

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