anonymous
  • anonymous
Evaluate the following limit: \[sqrt{(9+11x^{2})} / (6+3x)]\
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
as x -> inf and as x-> -(inf)
anonymous
  • anonymous
apply La Hopitals rule
anonymous
  • anonymous
differentiate top, differentiate the bottom, and try again

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anonymous
  • anonymous
y= ( 9 + 11x^2)^1/2 dy/dx = (1/2) ( 9 + 11x^2 )^-1/2 * 22x
anonymous
  • anonymous
= 11x / ( sqrt ( 9+11x^2) )
anonymous
  • anonymous
so we want to take the limit of\[\frac{ \frac{11x}{\sqrt{9+11x^2} } } { 3}\]
anonymous
  • anonymous
yea, that's what i got so far
anonymous
  • anonymous
applying the rule again because it is still indeterminate form "infinity/infinity"
anonymous
  • anonymous
so the (1/3) factor , bring that out then we have 11 / [ 11x / sqrt(9+11x^2) ]
anonymous
  • anonymous
so that gives what sqrt(9+11x^2) / x I think
anonymous
  • anonymous
\[\frac{\sqrt{9+11x^2}}{x}\]
anonymous
  • anonymous
sstill indeterminate , but if we apply the rule once more that x on the denominator will vanish
anonymous
  • anonymous
yeh I think something possibly went wrong, because it we differentiate that then we get back to 11x/ sqrt (9+11x^2)
anonymous
  • anonymous
I would just go with dominate term analysis here I think
anonymous
  • anonymous
I would just go with dominate term analysis here I think
anonymous
  • anonymous
when i apply the lospitl rule i don't have to do a quotien rule between the two func tion right?
anonymous
  • anonymous
i got the answers right thou thanks to inf is sqrt11/3 and to -(inf) is -sqrt11/3
anonymous
  • anonymous
dominate term up the top is \[x \sqrt{11}\] the dominate term on the bottom is x ( obviously )
anonymous
  • anonymous
so the quotient of these gives the answers sqrt(11) , and then you remember the factor of (1/3) we had earlier
anonymous
  • anonymous
how is it \[x sqrt {11}\]
anonymous
  • anonymous
do i just forget about the 9 ?
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty}\frac{\sqrt{9+11x^{2}}}{6+3x}\] \[ = \lim_{x \rightarrow \infty} \frac{x\sqrt{\frac{9}{x^2}+11}}{x(\frac{6}{x}+3)}\]
anonymous
  • anonymous
Becomes pretty easy from there.
anonymous
  • anonymous
Looks like \[\frac{\sqrt{11}}{3}\]
anonymous
  • anonymous
Did you follow that?
anonymous
  • anonymous
i didn't get how yu got sqrt11/3 from that. Did you just ignore everything that had an x ?
anonymous
  • anonymous
No. I took the limit as x goes to infinity. \(9/x^2\) will go to 0. 6/x will go to 0. The only things left will be the sqrt{11} and the 3 in the denominator.

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