Evaluate the following limit:
\[sqrt{(9+11x^{2})} / (6+3x)]\

- anonymous

Evaluate the following limit:
\[sqrt{(9+11x^{2})} / (6+3x)]\

- jamiebookeater

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- anonymous

as x -> inf and as x-> -(inf)

- anonymous

apply La Hopitals rule

- anonymous

differentiate top, differentiate the bottom, and try again

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## More answers

- anonymous

y= ( 9 + 11x^2)^1/2
dy/dx = (1/2) ( 9 + 11x^2 )^-1/2 * 22x

- anonymous

= 11x / ( sqrt ( 9+11x^2) )

- anonymous

so we want to take the limit of\[\frac{ \frac{11x}{\sqrt{9+11x^2} } } { 3}\]

- anonymous

yea, that's what i got so far

- anonymous

applying the rule again because it is still indeterminate form "infinity/infinity"

- anonymous

so the (1/3) factor , bring that out
then we have
11 / [ 11x / sqrt(9+11x^2) ]

- anonymous

so that gives what
sqrt(9+11x^2) / x I think

- anonymous

\[\frac{\sqrt{9+11x^2}}{x}\]

- anonymous

sstill indeterminate , but if we apply the rule once more that x on the denominator will vanish

- anonymous

yeh I think something possibly went wrong, because it we differentiate that then we get back to
11x/ sqrt (9+11x^2)

- anonymous

I would just go with dominate term analysis here I think

- anonymous

I would just go with dominate term analysis here I think

- anonymous

when i apply the lospitl rule i don't have to do a quotien rule between the two func tion right?

- anonymous

i got the answers right thou thanks
to inf is sqrt11/3 and to -(inf) is -sqrt11/3

- anonymous

dominate term up the top is \[x \sqrt{11}\]
the dominate term on the bottom is x ( obviously )

- anonymous

so the quotient of these gives the answers sqrt(11) , and then you remember the factor of (1/3) we had earlier

- anonymous

how is it \[x sqrt {11}\]

- anonymous

do i just forget about the 9 ?

- 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

Becomes pretty easy from there.

- anonymous

Looks like \[\frac{\sqrt{11}}{3}\]

- anonymous

Did you follow that?

- anonymous

i didn't get how yu got sqrt11/3 from that. Did you just ignore everything that had an x ?

- 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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