# limit as goes to infinity… I thought it is simple

I have to calculate the Asymptotes in the infinity(and minus infinity) for this function:

$f(x)=((x-7)(x+4))^{1/2}$

I know that

$\lim_{x \to\infty} f(x)/x= 1$

And I get into trouble with:

$\lim_{x \to\infty} f(x)-x$

which is $\lim_{x \to\infty} ((x-7)(x+4))^{1/2}-x = \lim_{x \to\infty} \sqrt{x^2-3x-28}-x$

wolfram alpha says it is $-3/2$ but i don't get why.... please help me with that, thanks

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"I know that $\lim_{x\to\infty} f(x)=1$." Then you know wrong, or there is something wrong with your definition of $f(x)$. – Thomas Andrews Jun 26 '14 at 13:23
right - I wrote here part of the calculation of a (i needed to divide by x) - editing – Yoav R. Jun 26 '14 at 13:26

The limit $$\lim_{x\to\infty} ((x−7)(x+4))^{1/2}$$ does not go to one. We can see this in a couple of ways. Let's take out $x^2$ from the radical, then we get: $$\lim_{x\to\infty} |x| \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)}$$

Then as $x \to \infty$ we see that the terms inside of the radical go to 1, and the $|x|$ goes to $+\infty$. Thus the limit diverges to $\infty$.

The same happens as $x \to -\infty$.

If you are trying to find the oblique asymptote, that is the long run behavior of $f$, then from what we have said so far, it looks like $f$ is tending to $x$ as $x\to\infty$. We can demonstrate this by writing:

$$\lim_{x\to\infty} \left(|x| \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - x\right)$$

$$=\lim_{x\to\infty} x\left( \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - 1\right)$$

$$=\lim_{x\to\infty} x\left( \frac{ \left(1-\frac7x\right)\left(1+\frac4x\right) - 1}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$

$$=\lim_{x \to\infty} \frac1x\left( \frac{ \left(x-7\right)\left(x+4\right) - x^2}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$

$$=\lim_{x \to\infty} \frac1x\left( \frac{ x^2 -3x - 28 - x^2}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$

$$=\lim_{x \to\infty} \left( \frac{ -3 - \frac{28}x}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$

$$=-3/2$$

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I know, I edited my question, I just forgot to divide by x – Yoav R. Jun 26 '14 at 13:27
Here I multiplied and divided by the conjugate and also multiplied and divided by $x^2$. Looking back at it, this could be done in fewer steps. – Joel Jun 26 '14 at 13:33
I don't understand step 2 to 3. Can you explain what you did there? Also, you found the limit to be $-\frac{3}{2}$, but what is the equation of the oblique asymptote? What does the $-\frac{3}{2}$ represent, the slope of the oblique asymptote? – rae306 Jun 26 '14 at 13:46
Step 2 to 3 involves multiplying and dividing by the conjugate. This is a multiplication by 1, so it doesn't change the value, but you can transform the equation that way. $$\sqrt{x}-1 \cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} = \frac{(\sqrt{x} - 1)(\sqrt{x}+1)}{\sqrt{x}+1} = \frac{x-1}{\sqrt{x}+1}$$ – Joel Jun 26 '14 at 14:36
Since we see that $\lim_{x\to \infty} f(x) - x = -3/2$ this tells us that $f(x)$ behaves like $x-3/2$ when $x$ is very large. – Joel Jun 26 '14 at 14:36

$$\sqrt{x^2-3x-28} - x = \frac{(\sqrt{x^2-3x-28} - x)(\sqrt{x^2-3x-28} + x)}{(\sqrt{x^2-3x-28} + x)}\\ = \frac{x^2-3x-28-x^2}{(\sqrt{x^2-3x-28} + x)} = \frac{-3x-28}{x(\sqrt{1-3/x-28/x^2} + 1)} \to \frac {-3}{2}$$

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Let me start from what Joel answered. We are concerned by the behavior of $$x\left( \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - 1\right)$$ for large values of $x$. Let us rewrite it as $$x\left(\sqrt{ 1-\frac7x} \sqrt{ 1+\frac4x}-1\right)$$ and now use $$\sqrt{ 1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Apply it for each radical and replace $y$ by $-\frac{7}{x}$ in the first and by $\frac{4}{x}$ in the second; now develop to get $$x\left(1-\frac{3}{2 x}-\frac{121}{8 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)-1\right)$$ and finally obtain $$x\left( \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - 1\right)=-\frac{3}{2}-\frac{121}{8 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$ So, you have the asymptote and moreover how the curve is with respect to it.

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$$\lim_{x \to\infty} \sqrt{x^2-3x-28}-x=\lim_{x \to\infty} x\Big(\sqrt{1-(\frac3x+\frac{28}{x^2})}-1\Big).$$ It is enough to limit the development of $\sqrt{1-t}=1-\frac 12t+O(t^2)$ to the term in $\frac1x$. $$\lim_{x \to\infty} x\Big(1-\frac3{2x}+O(\frac1{x^2})-1\Big)=-\frac32.$$

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