This is my third example of how to find the asymptotes of a function,
given its equation.
It is significantly harder than the previous two examples.
I will work with the function h, defined by: h(x) = x⋅arctan(x).
This function has TWO DIFFERENT slant asymptotes:
one as x approaches ∞, and one as x approaches -∞.
Interesting. I want to find those two asymptotes.
At this moment, I invite you to pause the video,
and try to solve the problem by any means you can.
And after that, continue watching.
DISCLAIMER. To be fair,
there is a faster way to solve this problem than what I am about to do.
There is an algorithm with a couple of formulas, which is more systematic.
However, what I want to show in this video is that,
even if nobody has taught me the method,
as long as I understand the definition,
if I am willing to try things and make errors, I can still figure things out.
I want to study the behaviour of this function as x approaches ∞.
The unusual part is arctan, (I know what to do with x)
but I remember that arctan by itself does have a horizontal asymptote.
I know the lim_(x → ∞) arctan(x) = π/2, so this suggests,
that as x is going towards ∞,
my function perhaps behaves like (π/2)⋅x,
because it's x times something with limit π/2.
So it suggests that my function is like a line: (π/2)⋅x.
Perhaps that could be an asymptote?
I'm not sure yet, but for now it's a conjecture (it's my candidate).
Well I don't know if it's going to work, but let's try to prove it!
For it to be an asymptote, I will need the following condition:
I would need the difference between the function and the asymptote,
(well the potential asymptote) to have limit 0 at ∞.
I don't know if this is going to work, but it's the only idea I have so far,
so let's try to prove it!
I'm trying to compute this limit.
As I have written it now, it is an indeterminate form,
because the limit of this term is ∞, and the limit of this term is also ∞,
so I have ∞ - ∞.
Well I notice I can take x as a common factor.
That's a good idea, because I have more tools for products than for differences.
Now I have the product of two things.
The limit of the first term of x, is ∞.
The limit of the second term has limit (π/2 - π/2): that's 0!
Okay, that's also an indeterminate form, but it's a product,
and I know I can transform every product into a quotient.
I simply wrote x, as 1/(1/x).
Now let's see what we have!
The numerator had limit 0 before, (it still has it)
and the denominator, 1/x, as x goes to ∞, has limit 0.
Perfect! This is 0/0,
so these are the conditions under which I can try to use L'Hôpital's Rule.
So I'm going to go ahead and use L'Hôpital's Rule.
This limit will be equal to... (assuming the new limit exists)
I have to take the derivative of the numerator:
the derivative of arctan(x) is 1/(1+x²).
The derivative of a constant is 0.
And the derivative of 1/x is -1/x².
And now let me regroup these things: (The arctan is gone; that's good news!)
and I am left with a much simpler limit.
So perhaps you know what the answer is:
it's just the leading coefficients,
or you want to factor x² from the numerator and denominator,
or you want to use L'Hôpital twice more...
But either way, after you do it,
we conclude that this limit is -1, which is great!
I computed the limit. It exists, so the original limit is -1.
But I wanted limit 0, so maybe it's not great.
So let me go back to the problem I was trying to solve.
I was wondering whether that limit was 0, but it isn't, it is -1.
So, my candidate for an asymptote is NOT an asymptote.
But not all is lost, because notice what I can do now:
if I add 1, on the left and on the right, I do get limit 0.
So it turns out,
that my function is not getting close to JUST (π/2)⋅x, but shifted by 1.
Or in other words, I can rewrite this as follows:
and I have proven that the limit as x goes to ∞,
of my original function - ((π/2)⋅x - 1), is 0.
In other words, the vertical distance between the graph of my function,
and the graph of this line, approaches 0 as x goes to ∞.
By definition, I have found an asymptote.
Well that was unexpected.
I couldn't have guessed from this equation...
...that this is the line that was going to appear.
I warned that this problem was more complicated.
Now, this function has a SECOND slant asymptote as x approaches -∞.
The process is similar, but the result is different.
I'm going to leave that as an exercise.
I invite you to try to do the same thing as x approaches -∞,
and you will get a different line.
So, this is a function with TWO DIFFERENT slant asymptotes.