Question

Find the oblique asymptote(s) of \[y=\sqrt{1+x^2}-4x\]

Answer 1

\[y=x \sqrt{\frac{1}{x^2}+1}-4x=x \left( \sqrt{\frac{1}{x^2}+1}-4 \right)\]To get the slope of the asymptotes you need to make x go to infinity for the second term, to get an equation of the form of a line. Then:\[\lim_{x \rightarrow \infty}\sqrt{\frac{1}{x^2}+1}-4=-3\]and\[\lim_{x \rightarrow \infty}-\sqrt{\frac{1}{x^2}+1}-4=-5\]
Therefore the aymptotes are -3x and -5x

Answer 2

I'm sorry, would you mind telling me how you come up with this way?

Answer 3

The asymptote is a line, therefore it obeys this equation: f=ax+b
I just put x out, and whats left is a and b, in this case there was no b.
But since the equation only aproaches this line on the infinity, you need to take the limit of the a you get for both signals of the square root.
Could you understand it this way?

Answer 4

Not fully, I'm sorry! I don't quite understand the part ''But since the equation only aproaches this line on the infinity, you need to take the limit of the a''

Answer 5

Your equation is clearly not a line, but it aproaches one when you get very far from the origin.
That equation of the line(ax+b) assumes that a and b are constants, otherwise this is not a line.
Since your equation is not a line but it aproaches one at the infinity, and the a you get(sqrt(1/x^2+1)-4) is a function of x, you need to take the limit when x goes to the infinity to find to wich slope this fuction aproaches.
If you take a closer look, this a is nothing but the derivative of the function y, wich is the slope at a certain point, the same logic applies to that, the slope will get more and more close to the slope of the line as you aproach infinity, thats why you need to take the limit.

Answer 6

This way of thinking about derivatives is probably the way you learned or will learn eventually, but I think logic is easier in this case.

Answer 7

''Since your equation is not a line but it aproaches one at the infinity, and the a you get(sqrt(1/x^2+1)-4) is a function of x''
Do you mean it tends to be a line when it approaches to infinity?

Answer 8

Yes, thats what an asymptote is, is the line to wich the function tends to. Make a graph and you'll see it.

Answer 9

And the line you meant is a straight line, I suppose?

Answer 10

Yes, always, thats the definition of asymptote.

Answer 11

I start getting your idea. Thanks!!

Answer 12

Your welcome. Here is a grph of your function close to 0.

Answer 13

And here when it gets really far from 0.

Answer 14

Thanks!!! It's very nice of you to attach the graphs for me!