The answer is (√3)/5.
I see two valid ways to go about this. One is faster and more intuitive for someone with experience, but the necessary justification is probably slightly too technical for a calculus class going over limits. The other is a lot clunkier, but it gets the job done with steps that are likely to be acceptable in this setting.
The fast method is as follows:
For a polynomial within a limit expression, if the limit runs as the variable goes to infinity (or negative infinity), it is typically safe to "drop" all of the terms in the polynomial that are not of the highest order. There are important caveats to this, which I can explain if asked, but it certainly applies to this problem. Thus, the limit can be reduced to
lim (x->∞) √(3(x^2))/(5x).
Now, for positive x, the numerator of the new limit expression is just (√3)x, leaving
lim (x->∞) ((√3)x)/(5x),
a rational function that reduces (for all nonzero x) to
lim (x->∞) (√3)/5
=(√3)/5.
Evaluating this limit using standard, well-justified steps takes a bit longer, but you can usually arrive at an answer this way and use it as a sanity check---or you can think about how to fill in the gaps between the initial expression and the "reduced" one. You can do this with "multiply-by-one" tricks, such as the following:
Note that
lim (x->∞) (5x)/(5x-9) = lim (x->∞) (1+(9/(5x-9)) = 1+0 = 1.
Also, note that lim (x->∞) √(3(x^2))/(5x)=(√3)/5.
Since multiplication is continuous for factors around 1 and (√3)/5, this means that the limit of the product of the above two limit expressions is equal to the product of limits:
lim (x->∞) ((5x)/(5x-9))*(√(3(x^2))/(5x))=1*(√3)/5=(√3)/5;
thus,
lim (x->∞) √(3(x^2))/(5x-9)=(√3)/5.
The same type of argument (with a little bit of added complexity from the radicals) would then be able to relate the numerator of that last limit expression to the numerator of the limit expression given in the original problem.
There is a complex bit of intuition involved with the polynomial trick, and it has everything to do with that limit of (5x)/(5x-9) and the steps around it, especially the "multiplying by one" trick. Searching for "leading terms" in expressions under limits is an invaluable skill, and it renders a large class of problems immediately solvable---this problem, for example, could be done almost instantly and entirely mentally. A more detailed and rigorous treatment of the material does not belong here, but I would be happy to go through some of the theory upon request.
I hope this helped!
