Confused about finding limit..

[quasar987]

I am asked to evaluate this limit

\lim_{x\rightarrow \infty}\frac{\sqrt{3x^2+11x+7}+\sqrt{3x^2-11x+7}}{x+\sqrt{x^2+7}}

but without use of the basic operations on limits (ex: the limit of a sum is the sum of the limits if they exist, etc.) (because they are to be introduced only in the next chapter). I know the limit is very easily computed: just divide the numerator and the denominator by x and the answer is \sqrt{3}:

= \lim_{x\rightarrow \infty}\frac{\sqrt{3+\frac{11}{x}+\frac{7}{x^2}}+\ sqrt{3-\frac{11}{x}+\frac{7}{x^2}}}{1+\sqrt{1+\frac{7}{x^ 2}}} = \sqrt{3}

But what is the "method" to arrive to this conclusion? The facts that

\lim_{x\rightarrow \infty}\frac{11}{x} = 0

and that

\lim_{x\rightarrow \infty}\frac{7}{x^2} = 0

are useless if we cannot use the operations on limits. I hope I made clear where my confusion comes from. Thanks.

[kreil]

There are 3 rules of infinite limits:

1. If the power of the top is the same as the power of the bottom, take the coefficients:


\lim_{x\rightarrow \infty} \frac {4x^2 + 3x + 7}{2x^2 + 2x + 1} = \frac {4}{2}=2


2. If the power of the bottom is larger than the power of the top, the limit is zero:


\lim_{x\rightarrow \infty} \frac {x^2 + 4x}{x^3 + 7x} = 0



3. If the power of the top is larger than the power of the bottom, the limit does not exist:


\lim_{x\rightarrow \infty} \frac {x^3 + 3x - 7}{x^2 + 4x} = dne


I don't think these are operations; they are just facts. In your case:


\lim_{x\rightarrow \infty}\frac{\sqrt{3x^2+11x+7}+\sqrt{3x^2-11x+7}}{x+\sqrt{x^2+7}}


the top and bottom are only to the power x^1 and so you take the coefficients, \frac {\sqrt{3}}{1}=\sqrt{3} Hopefully this helps you.

[quasar987]

I'm afraid I'm not allowed to use your rules. Only the definition of limit, the caracterisation with sequences and a few theorems about the unicity and the "limit to the right, limit to the left" caracterisation.

Let me ask the question another way: What are the steps between

\lim_{x\rightarrow \infty}\frac{\sqrt{3+\frac{11}{x}+\frac{7}{x^2}}+\ sqrt{3-\frac{11}{x}+\frac{7}{x^2}}}{1+\sqrt{1+\frac{7}{x^ 2}}}

and

= \sqrt{3}

I tought of something this morning, could it be it? Now that we suspect using our "sense" that the limit is \sqrt{3}, we may use the \epsilon-\delta definition to prove that it is indeed \sqrt{3}. But I don't really see how. :shy:

[sal]

I'm afraid I'm not allowed to use your rules. Only the definition of limit, the caracterisation with sequences and a few theorems about the unicity and the "limit to the right, limit to the left" caracterisation.

Let me ask the question another way: What are the steps between

\lim_{x\rightarrow \infty}\frac{\sqrt{3+\frac{11}{x}+\frac{7}{x^2}}+\ sqrt{3-\frac{11}{x}+\frac{7}{x^2}}}{1+\sqrt{1+\frac{7}{x^ 2}}}

and

= \sqrt{3}

I tought of something this morning, could it be it? Now that we suspect using our "sense" that the limit is \sqrt{3}, we may use the \epsilon-\delta definition to prove that it is indeed \sqrt{3}. But I don't really see how. :shy:
Right, but because you're taking the limit as x goes to infinity, not zero, I would tend to use "N" rather than epsilon.

To save space, let

f(x) = \frac{\sqrt{3+\frac{11}{x}+\frac{7}{x^2}}+
\sqrt{3-\frac{11}{x}+\frac{7}{x^2}}}
{1+\sqrt{1+\frac{7}{x^ 2}}}


To apply the definition directly, what you need to do is assume you have a value for delta, and then solve for N in the equation:

f(N) - \sqrt{3} = \delta

And once you've done that, you need to show that

x > N \Rightarrow |f(x) - \sqrt{3}| < |f(N) - \sqrt{3}|

It looks likely to get messy, which, if nothing else, illustrates why we don't generally work directly from the definition of the limit.

[quasar987]

Great, thanks.