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

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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

Mathematics
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Is it \[\lim_{x \rightarrow \infty} \frac{4}{9x+3}\] for your question?
yes sorry!! I was busy doing other practice questions for my test tomorrow.
Never mind :) 1. Divide both numerator and denominator by x, what do you get?

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|dw:1350532389029:dw|
Yup! Now, you need to know that \[\lim_{x \rightarrow \infty}\frac{1}{x} = 0\]Now, can you evaluate the limit?
what does the 1/x = 0 mean? does that also mean that 4/x - 0?
4/x=0
It's \[\lim_{x \rightarrow \infty \frac{1}{x}} = 0\] Imagine 1 is divided by a very large number, then you probably get 0.
Aww.. Bad typing! It's \[\lim_{x \rightarrow \infty} \frac{1}{x}=0\]
so 4 divide by a very large number is pretty much 0, so I think the answer should be 0/9.
Yup! Which is...?
0
You got it!
:) Thank you! It's greatly appreciated!
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 \[\lim_{x \rightarrow \infty}\frac{1}{x} = 0\]
One question, how did you know to divide by x at the first?

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