## ConDawg Group Title Evaluate the following limit. If the answer is positive infinite, type "I"; if negative infinite, type "N"; and if it does not exist, type "D". lim ---> 4/9x+3 x goes to infinity one year ago one year ago

1. Callisto Group Title

Is it $\lim_{x \rightarrow \infty} \frac{4}{9x+3}$ for your question?

2. ConDawg Group Title

yes sorry!! I was busy doing other practice questions for my test tomorrow.

3. Callisto Group Title

Never mind :) 1. Divide both numerator and denominator by x, what do you get?

4. ConDawg Group Title

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5. Callisto Group Title

Yup! Now, you need to know that $\lim_{x \rightarrow \infty}\frac{1}{x} = 0$Now, can you evaluate the limit?

6. ConDawg Group Title

what does the 1/x = 0 mean? does that also mean that 4/x - 0?

7. ConDawg Group Title

4/x=0

8. Callisto Group Title

It's $\lim_{x \rightarrow \infty \frac{1}{x}} = 0$ Imagine 1 is divided by a very large number, then you probably get 0.

9. Callisto Group Title

Aww.. Bad typing! It's $\lim_{x \rightarrow \infty} \frac{1}{x}=0$

10. ConDawg Group Title

so 4 divide by a very large number is pretty much 0, so I think the answer should be 0/9.

11. Callisto Group Title

Yup! Which is...?

12. ConDawg Group Title

0

13. Callisto Group Title

You got it!

14. ConDawg Group Title

:) Thank you! It's greatly appreciated!

15. Callisto Group Title

Just a little summary of the above: $\lim_{x \rightarrow \infty} \frac{4}{9x+3}$Divide both numerator and denominator by x$\lim_{x \rightarrow \infty} \frac{\frac{4}{x}}{\frac{9x+3}{x}}$$=\lim_{x \rightarrow \infty} \frac{\frac{4}{x}}{9+\frac{3}{x}}$$= \frac{\lim_{x \rightarrow \infty}(\frac{4}{x})}{\lim_{x \rightarrow \infty}(9)+\lim_{x \rightarrow \infty}(\frac{3}{x})}$And then evaluate the limit by using $\lim_{x \rightarrow \infty}\frac{1}{x} = 0$

16. ConDawg Group Title

One question, how did you know to divide by x at the first?