wesdg1978
  • wesdg1978
lim x-->inf sqrt(x^2+18x) -x How do you find the limit?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Jhannybean
  • Jhannybean
\[\large \lim_{x \rightarrow \infty}\sqrt{x^2+18x}-x\] Something to do with conjugates perhaps?
Jhannybean
  • Jhannybean
\[\large \lim_{x \rightarrow \infty }(\sqrt{x^2+18x}-x)\cdot \frac{(\sqrt{x^2+18x}+x)}{(\sqrt{x^2+18x}+x)}\]\[\large \lim_{x \rightarrow \infty } \frac{(x^{2}+18x-x^2)}{\sqrt{x^2 +18x}+x}\]\[\large \lim_{x \rightarrow \infty } \frac{18x}{\sqrt{x^2 +18x}+x}\]Divide by highest power in denominator. \[\large \lim_{x \rightarrow \infty } \frac{18x/x}{\sqrt{x^2/x^2 +18x/x^2}+x/x}\]\[\large \lim_{x \rightarrow \infty } \frac{18}{\sqrt{1 +0}+1}= \frac{18}{2} = 9\]
Jhannybean
  • Jhannybean
First turn it into a fraction so you can cancel stuff under a square root. Upon making it a fraction you divide by the highest power in the denominator which you simplify to find your answer.

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Jhannybean
  • Jhannybean
And remember, depending on which limit you are taking (either from the left, or the right) you would either be dividing by +x or -x. And in dealing with the square root, you would be dividing by the highest power that's inside the square root,which is x^2.
wesdg1978
  • wesdg1978
How does 18x/x^2 go to zero?
Jhannybean
  • Jhannybean
Remember, you're not dividing by x^2 in the numerator. you're dividing by the highest power in the denominator, which is \(\large \sqrt{x^2} = \left|x\right| = +x \) sine you're tending to positive infinity.
Jhannybean
  • Jhannybean
since*
wesdg1978
  • wesdg1978
In the denominator, 18x/x^2 goes to zero, I don't understand how that works
Jhannybean
  • Jhannybean
Well, what happens when you divide x/x^2? What do you get?
wesdg1978
  • wesdg1978
1/x
wesdg1978
  • wesdg1978
the opposite of x^2/x
Jhannybean
  • Jhannybean
Ok. and so if we multiplied it by 18 we would have 18/x, correct?
Jhannybean
  • Jhannybean
as x approaches infinity, 18/x -> 0 because any constant divided by a large number like infinity will tend to 0.
wesdg1978
  • wesdg1978
Oh, ok, I get it now, the larger x gets, the smaller the number gets towards 0.
wesdg1978
  • wesdg1978
Thanks for your help!
Jhannybean
  • Jhannybean
No problem :) glad you understand this!
wesdg1978
  • wesdg1978
I'm trying to understand
Jhannybean
  • Jhannybean
Whats confusing you?
wesdg1978
  • wesdg1978
I was just saying in general, as it applies to calculus, that I'm trying to understand.
Jhannybean
  • Jhannybean
Well, it helps you determine asymptotes...helps determine the behavior of the function coming in from the left and right...etc.
wesdg1978
  • wesdg1978
I appreciate your help
Jhannybean
  • Jhannybean
no problem :)

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