Evaluate the limit lim (√(9+4x^(2)))/(11+10x) x->INF
\[\lim_{x\to\infty}{\sqrt{9+4x^2}\over11+10x^2}\]correct?
(11+10x)
\[\lim_{x\to\infty}{\sqrt{9+4x^2}\over11+10x}\]
yup
much easier... divide top and bottom by x, then take the limit
so its √(9+4x^)/(11+10)?
√(9+4x)/(11+10)?
you must be more careful with the numerator
what is\[{\sqrt{9+4x^2}\over x}\]simplified?
9+4x?
1) where did the square root go? 2) how does the square root sign affect trying to divide by x?
the answers: 1) the square root must stay 2) in order to divide this by x that means we must square it, to get it into the radical
square the numerator
?
9+4x^2
you have to square x to put it into the radical, just as it works the other way....
\[\sqrt{\frac a{x^2}}=\frac1x\sqrt{a}\]so it works the other way too
\[{\sqrt a\over x}=\sqrt {a\over{x^2}}\]
so we have\[{\sqrt{9+4x^2}\over x}=\sqrt{{9+4x^2\over x^2}}=\sqrt{\frac9{x^2}+4}\]for the numerator
why did you square x?
it's perhaps best seen by example...
\[{\sqrt {16}\over2}=\sqrt{\frac{16}{2^2}}=\sqrt{\frac{16}4}=2\]\[{\sqrt {25}\over5}=\sqrt{\frac{25}{5^2}}=\sqrt{\frac{16}4}=1\]
etc.
try it with uglier numbers if you like, I don't want to type it all out
in principle it works the same as simplifying\[\sqrt{24}\]if you know how to do that:\[\sqrt{24}=\sqrt4\cdot6=2\sqrt6\]
only here we are doing it with division
\[\sqrt{\frac{50}4}=\frac{\sqrt{50}}2=\frac{2\sqrt5}2=\sqrt5 \]
I'm sorry, but I'm a bit too tired to go into a more detailed explanation. I hope you can meditate on this and decipher it.
yup i got it thank you
well what I wrote above was wrong, should have been:\[\sqrt{\frac{50}4}=\frac{\sqrt{50}}2=\frac{5\sqrt2}2\]told you I'm tired. later!
hhhh thanks
2/10 got it
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