Finding a limit

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Violagirl

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Hi, I am really confused as to how to solve this problem. I got an answer of six but in my book, it says that the answer is -oo. Here is what I did:

1. lim 6x^2/x+3= 6x^2/6x^2/x/x+3/x^2=6/1+0=6. If this is not correct, how would you do the problem? And if you could explain it too, that would be great!
-->-oo Thanks!
 
Violagirl said:
Hi, I am really confused as to how to solve this problem. I got an answer of six but in my book, it says that the answer is -oo. Here is what I did:

1. lim 6x^2/x+3= 6x^2/6x^2/x/x+3/x^2=6/1+0=6. If this is not correct, how would you do the problem? And if you could explain it too, that would be great!
-->-oo Thanks!
Click to expand...

What in the ...?

If the power is higher in the numerator and x is tending to +/- infinity, the limit is never a finite number.
 
The general case can be read in any pre-calculus book (coded as "finding a horizontal asymptote"), or calculus book.

\(\displaystyle \frac{6x^2}{x+3}\) is negative when x<-3. As x gets farther to the left from -3 on the real axis, what happens to the value of the rational function?


If you'd like, assuming x is somewhere down in negative land, simplify it to: \(\displaystyle \frac{6x}{1 + \frac{3}{x}}\)
 
Ok gotcha! So in general, when you have an X value with a degree and a regular number like that, would you divide by the degree of x that is with it in either the numerator or denominator?
 
That may help, but for rational functions there are simpler rules.

You could have also rewritten it as:

\(\displaystyle \frac{6}{\frac{1}{x} + \frac{3}{x^2}}\), but I felt it made it "harder" to see the function's value becomes negative beyond -3.

In general, given x is approaching either positive or negative infinity: if the degree is higher in the numerator, the function will approach either positive or negative infinity. If they are the same, the function will tend to a real number. If the degree in the top is smaller, the function will tend to zero.
 
[attachment=0:3bfjjkkv]qqq.jpg[/attachment:3bfjjkkv]

Observed the above graph of 6x^2/(x+3).

Domain: (-?,-3)U(-3,?)

Range: (-?,-72]U{0,?)

Note x = -3 is a vertical asymptote and y = 6x is an oblique asymptote.

From this information, one should be able to find the limit as x goes to ± ?.
 

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