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Limits
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\[\lim_{x \rightarrow -\infty}\frac{ 4x }{ \sqrt{3+x^2} }\]
Have you learnt the l'hopital rule?
We might have gone over it, but I can't recall it
Do it analytically is possible too. As x becomes very large, the 3 is insignificant and can be ignored. Simplify this:\[\Large \frac{ 4x }{ \sqrt{x^2} }\]
It'll simplify to:\[\Large \frac{ 4x }{ \left| x \right| }\]which means if it's x is negative...
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Alright, I have it up to here. Where do I go form here?
Just simplify. If you can't do it as is, imagine if x=-1, or something like that. You'll see it's very easy.
\[\frac{ 4(-1) }{ \left| -1 \right| }\] -4?
exactly
Alright, well thanks a lot!
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no problem
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