anonymous
  • anonymous
Evaluate the given function..............( see the comments for the question)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
here is the question
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anonymous
  • anonymous
Please explain how did you arrive at the answer, I don't know how to use L'Hopital's rule, nor has my professor taught us that.
anonymous
  • anonymous
so , I found 1/10

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anonymous
  • anonymous
Lemme try.....
anonymous
  • anonymous
Sorry, its the wrong answer
anonymous
  • anonymous
ok let me check my work .)
anonymous
  • anonymous
r u sure it's wrong .s
anonymous
  • anonymous
yeah, i did check it, the question system says, its wrong.
anonymous
  • anonymous
Okay, I double checked, your answer is right, sorry...
anonymous
  • anonymous
How did you get it??
anonymous
  • anonymous
Change y=1/x and y-->0 sqrt(25/y^2 + 1/y) - 5/y= sqrt[(25 + y)/y^2] - 5/y=[sqrt(25+y) - 5]/y Let's complete the square from 25+y by +/-(y/10)^2 25+y=25 + 2* 5* y/10 + (y/10)^2 - (y/10)^2=(5+y/10)^2-(y/10)^2 Continue sqrt[(5+y/10-5]/y=1/10 lim(x-->infinity)=1/10
anonymous
  • anonymous
okay why did we do the first step?
anonymous
  • anonymous
@Emineyy, why did u do the first step??
anonymous
  • anonymous
Dude, could u tell me how to approach solving this question??
DebbieG
  • DebbieG
Wait.... am I missing something? It's: \[\Large \lim_{x \rightarrow \infty}\sqrt{25x^2+x}-5x\] right? The limit of a square root is the square root of a limit. The limit of a polynomial as \({x \rightarrow \infty}\) is determined by the behavior of the leading term (sign of coefficient, and degree). The limit of a difference is the difference of the limits. I think that's all you need to use here??
anonymous
  • anonymous
yup, you got the question right.
anonymous
  • anonymous
the answer is correct
anonymous
  • anonymous
I checked it, initially the website said, its wrong, but when I double checked it came out to be right
DebbieG
  • DebbieG
It said 1/10 was the limit?
anonymous
  • anonymous
yes
DebbieG
  • DebbieG
Oh nvr mind... that's what Wolfram says too. OK, I'm bowing out, I'm obviously missing something here, lol.
anonymous
  • anonymous
:)))
anonymous
  • anonymous
Hey, you are back, help me out here bud
anonymous
  • anonymous
So, I don't get why we did the first step and why we did the last step where 5+y/10-5 = 1/10
anonymous
  • anonymous
@Emineyy
anonymous
  • anonymous
help me out bud....@Emineyy
anonymous
  • anonymous
I mean, isn't y = 1/x and if that's so, then shouldn't it be 0??
DebbieG
  • DebbieG
Well, I do see why it isn't as simple as I thought initially... duh.... The limit of the square root part is + infinity The limit of the -5x is - infinity So it isn't like they can just be added. So let me see if I can make sense of her approach... she's making a substitution, re-writing with \(y=1/x\) so then \(x=1/y\) and \(x^2=1/y^2\). \(\sqrt{25/y^2 + 1/y} - 5/y= \sqrt{(25 + y)/y^2} - 5/y=\dfrac{\sqrt{(25+y)} - 5}{y} \) =====I don't understand what the purpose of all this completing the square stuff was about...??===== Let's complete the square from 25+y \(25+y=25 + 2\cdot5\cdot \dfrac{y}{10} + (\dfrac{y}{10})^2 - (\dfrac{y}{10})^2=(5+y/10)^2-(y/10)^2 \) OK, so: \(25+y=(5+y/10)^2-(y/10)^2 \) ===================================== Now back to the limit... with the substitution, we have: \(\lim_{y \rightarrow 0}\dfrac{\sqrt{(25+y)} - 5}{y} \) which is a 0/0 form and L'hopital's rule applies. \(\dfrac{d}{dy}(25+y)^{1/2} - 5=\dfrac{1}{2}(25+y)^{-1/2}=\dfrac{1}{2}\dfrac{1}{(25+y)^{1/2}}\) so: \(\lim_{y \rightarrow 0}\dfrac{1}{2}\dfrac{1}{(25+y)^{1/2}}=\dfrac{1}{2\cdot5}=\dfrac{1}{10}\) (And the derivative of the "y" den'r is just 1).

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