anonymous
  • anonymous
Find the limit: lim as x approaches infinity 1-2x / sqrt(9x^2 +5)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) like this?
anonymous
  • anonymous
yes
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle-\lim_{x \rightarrow ~\infty }\frac{(2x-1)}{\sqrt{9x^2+5}}}\) now apply LHS

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SolomonZelman
  • SolomonZelman
I factored the top out of -1 and took the constant (the -1) out. as lim cf(x) = c lim f(x)
anonymous
  • anonymous
no not yet
SolomonZelman
  • SolomonZelman
oh, then disregard what I said, since you haven't learned the L'Hospital's rule
SolomonZelman
  • SolomonZelman
are you asked to do this in a particular way, oor you can use any method? I would simply use a numerical approach
SolomonZelman
  • SolomonZelman
(to plug in very large number to see what the limit is going to approach)
SolomonZelman
  • SolomonZelman
or you can go about is intuitively
anonymous
  • anonymous
we 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
  • SolomonZelman
intuitively: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) 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: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) (and it will be an equivalent)
SolomonZelman
  • SolomonZelman
oh, I meant to say, to write it as \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}\)
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}\) simplify the square root \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{3x}}\) x's cancel, and you get?
anonymous
  • anonymous
-2/3 ?
SolomonZelman
  • SolomonZelman
yup
SolomonZelman
  • SolomonZelman
this limit is -2/3
anonymous
  • anonymous
ohhhh! makes so much sense. I had this problem on my exam..... I got it wrong
anonymous
  • anonymous
thank you though for clarifying this that way I don't make the same mistake. Thank you.
SolomonZelman
  • SolomonZelman
Yw....

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