# Fnd the limit of sqrt(x^2 + 4x) - x as x goes to infinity

• dranseth
In summary: I am showing you now.You can think of the "conjugate factor" method as being based on the "differences of squares" formula, which is a special case of "differences of powers" (which is also a special case of "sums of powers"). This is a standard algebraic technique to use when you have differences of powers, and it can be generalized. The most important idea in using it, however, is to keep in mind that you are trying to get rid of differences between terms, and to see to it that they will cancel. This is what I did in the numerator of my work above. The denominator didn't have any differences, but I did do
dranseth

## Homework Statement

I'm a bit puzzled with a limit, as is my teacher. We came across a question that could be both factored or rationalized, but when factored, the answer is different. Perhaps someone has an explanation?

Here it is.

lim ((√x^2+4x)-x)
x->∞

I'm not going to rationalize it, because I know that is right. I'm going to show you how I'm factoring it, and perhaps someone can explain why it is not working. Thanks

lim ((√x^2+4x)-x)
x->∞

lim (x(√1+4/x)-x)
x->∞

lim x((√1+0)-1)
x->∞

lim x(0)
x->∞

lim = 0
x->∞

...

dranseth said:
lim (x(?1+4/x)-x)
x->?

lim x((?1+0)-1)
x->?

Here's the source of the difficulty. You are starting out with what is called an "indeterminate difference", because you are subtracting infinity from infinity, which can give you anything at all. By factoring out the "x", the other factor is now

(?1+4/x) - 1 ,

which, for the limit at infinity, is indeed zero. The trouble is that your full expression now gives a limit of infinity times zero, which is called an "indeterminate product", as this can also give you anything at all.

The most direct way of dealing with such indeterminate differences is to use the "conjugate factor" method (that's what I call it -- I don't know its "official" name, or if it even has one). Multiply your expression by the ratio

((?x^2+4x) + x) / ((?x^2+4x) + x) , which is just one. This gives you the quotient

[((?x^2+4x) - x) · ((?x^2+4x) + x)] / ((?x^2+4x) + x) .

The factors in the numerator are the factors of a difference of two squares, which is to say that if you multiply those two binomials, the "cross terms" will cancel. You now have

[(x^2+4x) - (x^2)] / ((?x^2+4x) + x) , which simplifies to

(4x) / ((?x^2+4x) + x) .

The point of this method is that it eliminates places where there are differences between terms, which lead to problems in taking limits. We still want the limit at infinity, so divide numerator and denominator here by x , which leads to

4 / ((?1 + 4/x) + 1) ,

for which the limit at infinity is 4 / [(?1+0) + 1] = 2 . Putting your original function into a graphing routine will confirm that it has a horizontal asymptote of y = 2.

EDIT: Either the board or my browser doesn't like the square root symbol you used when I put it in my reply. So, if you're seeing ?'s all over the place, those are supposed to be radicals...

Last edited:
lim (x(√1+4/x)-x)
x->∞

lim x((√1+0)-1)
x->∞

why did that 4/x become 0? you can't just plug the limit in one of the factor like that because what you've got there is an indeterminate form ∞0.

I was taught to plug a zero in wherever x is in the denominator. ie. 4/x = 0 when dealing with limits of infinity

is there any way to tell when you've got a function in indeterminate form before factoring?

dranseth said:
is there any way to tell when you've got a function in indeterminate form before factoring?

The first thing to do with a limit problem is just follow the "limit laws" and see what comes out. If you can follow proper rules of arithmetic to get an answer (including infinity), then that result is the limit. The following results are "indeterminate" and require some other approach to be used to evaluate the limit properly:

0/0 , $$\frac{\infty}{\infty}$$ (indeterminate ratios)
$$0 · \infty$$ (indeterminate product)
$$\infty - \infty$$ (indeterminate difference)
$$0^{0} , 1^{\infty}$$ (indeterminate powers)

I was taught to plug a zero in wherever x is in the denominator. ie. 4/x = 0 when dealing with limits of infinity

That is true by the "reciprocal powers" limit law: the limit at positive or negative infinity of a real number divided by a positive real power of x is zero. Where you had applied it is fine; the difficulty for that limit lay elsewhere. You have to look over the behavior of the terms in the function in the infinite limit to see if there will be trouble...

Last edited:

## 1. What is the limit of sqrt(x^2 + 4x) - x as x goes to infinity?

The limit of sqrt(x^2 + 4x) - x as x goes to infinity is infinity. This means that as x approaches infinity, the value of the expression also approaches infinity.

## 2. How do you find the limit of sqrt(x^2 + 4x) - x as x goes to infinity?

To find the limit of sqrt(x^2 + 4x) - x as x goes to infinity, you can use the rules of limits and algebraic manipulation. First, you can rewrite the expression as sqrt(x^2 + 4x) - x = sqrt(x^2(1 + 4/x)) - x = x * sqrt(1 + 4/x) - x. Then, you can use the limit rule that states that the limit of a product is equal to the product of the limits. Since the limit of x as x goes to infinity is infinity, the limit of the entire expression is also infinity.

## 3. Can the limit of sqrt(x^2 + 4x) - x as x goes to infinity be negative infinity?

No, the limit of sqrt(x^2 + 4x) - x as x goes to infinity cannot be negative infinity. This is because the expression contains a square root, which is always positive. As x approaches infinity, the value of the expression may become more and more negative, but it will never reach negative infinity.

## 4. Is it possible for the limit of sqrt(x^2 + 4x) - x as x goes to infinity to not exist?

Yes, it is possible for the limit of sqrt(x^2 + 4x) - x as x goes to infinity to not exist. This can happen if the expression has oscillations or jumps as x approaches infinity. In this case, the limit does not approach a single value and therefore does not exist.

## 5. Can you use L'Hopital's rule to find the limit of sqrt(x^2 + 4x) - x as x goes to infinity?

No, L'Hopital's rule cannot be used to find the limit of sqrt(x^2 + 4x) - x as x goes to infinity. This is because the expression does not have the form of a fraction, which is a requirement for L'Hopital's rule to be applicable.

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