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anonymous
 one year ago
Find the limit: lim as x approaches infinity 12x / sqrt(9x^2 +5)
anonymous
 one year ago
Find the limit: lim as x approaches infinity 12x / sqrt(9x^2 +5)

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{12x}{\sqrt{9x^2+5}}}\) like this?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{(2x1)}{\sqrt{9x^2+5}}}\) now apply LHS

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I factored the top out of 1 and took the constant (the 1) out. as lim cf(x) = c lim f(x)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, then disregard what I said, since you haven't learned the L'Hospital's rule

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1are you asked to do this in a particular way, oor you can use any method? I would simply use a numerical approach

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1(to plug in very large number to see what the limit is going to approach)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1or you can go about is intuitively

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we can use any method. I plugged in the equation into my calculator and everytime I'd plug in larger number it would get closer to .66636 or some value around that.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1intuitively: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{12x}{\sqrt{9x^2+5}}}\) the top of the limit as x goes to very large numbers is just 2x, and the +1 there becomes even less significant the greater x value we choose. the bottom of the limit as x goes to very large numbers is just √(9x²) , becuase the +5 at some point (just like +1 on top) becomes insignificant. So you can write it as: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{12x}{\sqrt{9x^2+5}}}\) (and it will be an equivalent)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, I meant to say, to write it as \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{2x}{\sqrt{9x^2}}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{2x}{\sqrt{9x^2}}}\) simplify the square root \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{2x}{3x}}\) x's cancel, and you get?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1this limit is 2/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhhh! makes so much sense. I had this problem on my exam..... I got it wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you though for clarifying this that way I don't make the same mistake. Thank you.
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