Justify that the limiting form of `V(x) = 350x/(1+x^2)^(3/4)` when x is large is `V(x) = 350/sqrt(x)`.

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txmedteach eNotes educator| Certified Educator

This is effectively saying "Calculate the expression you will evaluate when looking at the limit to infinity," which thankfully isn't too bad with this equation. Let's retype it below in a familiar format:

`V(x) = (350x)/(1+x^2)^(3/4)`

When we take the limit as `xtooo`, we see that one important thing happens in the denominator. I always called this the "Bill Gates" Rule (described slightly more formally at the link below) and it allows us to convert the above equation to the following when we take the limit as we approach infinity:

`lim_(x->oo)V(x) = lim_(x->oo) (350x)/(1+x^2)^(3/4) = lim_(x->oo)(350x)/(x^2)^(3/4)`

The idea is that when you have `x` approach infinity, `x^2`is so much bigger than 1 that we can effectively consider the expression in the denominator equivalent to `x^2`. For example, if Bill Gates were to square his $60 billion to make some ungodly amount of money, it wouldn't make a difference to him whether or not he got the extra dollar! Seriously, would you really notice a difference between $3,600,000,000,000,000,000,000 and $3,600,000,000,000,000,000,001? I certainly wouldn't.

Moving on, now that we know that as we have `x` become very large, we can consider V(x) to be effectively the following:

`lim_(x->oo)V(x) = lim_(x->oo)(350x)/(x^2)^(3/4)`

Now, we simply do some exponent simplification. Notice when you take a power of x to another power, you can simply multiply the exponents. In our case, we multiply 2 with 3/4 to get 3/2:

`= lim_(x->oo) (350x)/x^(3/2)`

Now, we're going to do SOME MORE exponent simplification. Recall, if we have `x^a/x^b`, then we can simplify by having the above expression be equivalent to `x^(a-b)`. That's what we'll do next:

`=lim_(x->oo)350x^(1-3/2) = lim_(x->oo)350x^(-1/2)`

Continuing, any time we have `x` taken to a negative power, we simply take the same power of `x` and throw it in the denominator:

`=lim_(x->oo) 350/x^(1/2)`

Now, recall that taking `x` to the 1/2 power is just another way of saying `sqrt(x)`, and we'll get the answer in the desired form to sum everything up:

`lim_(x->oo) V(x) = lim_(x->oo)350/sqrt(x)`

Now, we're going to stop here with the limit aspect of it, though, because if we were to continue to evaluate this limit, then our answer would be 0, because as `x` increases, we start to also completely outweigh the 350 in the numerator, not just the 1 in the denominator of the original expression!

I hope that helps!