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# limits -chapter help me please?

lim 3x^2-4x / 18x^2-5x^2 =infinity how?

x....infinity

Update:

if it is

limit (3x^2-4x) /(18x^6-5x^2)= infinity,how

x.....infinity

### 3 Answers

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- 1 decade agoFavorite Answer
hmm, your formula is not very clear:

if you mean:

(3x^2 - 4x) / (18x^2 - 5x^2) = (3x^2 - 4x) / 13x^2 = 3x^2 / 13x^2 - 4x/13x^2 = 3/13 - 4/13x = 3/13 (4/13x limits to 0)

if you mean:

(3x^2 - 4x) / 18x^2 - 5x^2 = 3x^2/18x^2 - 4x/18x^2 - 5x^2 = 3/18 - 4/18x - 5x^2 = -infinity

- 1 decade ago
The value of this limit is not infinity but is 3/13 apply L'Hospital's Rule

since it is of infinity by infinity form

- 1 decade ago
Hell no I will never help you out man. I don't roll like that Dog. peace out ganster

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