## anonymous 5 years ago Evaluate the following limit: $sqrt{(9+11x^{2})} / (6+3x)]\ • This Question is Closed 1. anonymous as x -> inf and as x-> -(inf) 2. anonymous apply La Hopitals rule 3. anonymous differentiate top, differentiate the bottom, and try again 4. anonymous y= ( 9 + 11x^2)^1/2 dy/dx = (1/2) ( 9 + 11x^2 )^-1/2 * 22x 5. anonymous = 11x / ( sqrt ( 9+11x^2) ) 6. anonymous so we want to take the limit of\[\frac{ \frac{11x}{\sqrt{9+11x^2} } } { 3}$

7. anonymous

yea, that's what i got so far

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

$\frac{\sqrt{9+11x^2}}{x}$

12. anonymous

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

13. anonymous

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

14. anonymous

I would just go with dominate term analysis here I think

15. anonymous

I would just go with dominate term analysis here I think

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

how is it $x sqrt {11}$

21. anonymous

do i just forget about the 9 ?

22. anonymous

$\lim_{x \rightarrow \infty}\frac{\sqrt{9+11x^{2}}}{6+3x}$ $= \lim_{x \rightarrow \infty} \frac{x\sqrt{\frac{9}{x^2}+11}}{x(\frac{6}{x}+3)}$

23. anonymous

Becomes pretty easy from there.

24. anonymous

Looks like $\frac{\sqrt{11}}{3}$

25. anonymous

26. anonymous

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

27. 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.