Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
"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 at 13:23
1  
right - I wrote here part of the calculation of a (i needed to divide by x) - editing –  Yoav R. Jun 26 at 13:26

4 Answers 4

up vote 2 down vote accepted

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:

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

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

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

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

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

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

$$=-3/2$$

share|improve this answer
1  
I know, I edited my question, I just forgot to divide by x –  Yoav R. Jun 26 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 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? –  Rainier van Es Jun 26 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 at 14:36
1  
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 at 14:36

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

share|improve this answer

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.