If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 1

Lesson 19: Unbounded limits (vertical asymptotes)

# Analyzing unbounded limits: mixed function

AP.CALC:
LIM‑2 (EU)
,
LIM‑2.D (LO)
,
LIM‑2.D.1 (EK)
,
LIM‑2.D.2 (EK)
Sal analyzes the behavior of f(x)=x/[1-cos(x-2)] around its asymptote at x=2.

## Want to join the conversation?

• what happens when you're given f(x)= 3e^-x + 2 ? how do you determine the limit at infinity ?
• 3e^(-x)=3/(e^x). So as x gets very large, e^x gets incredibly large, and so 3/(e^x) goes to 0.

So you're just left with 2.
• How would you do that sort of problem without a calculator? Is there any algebra you could use?
• When I did this problem, I did not use any algebra per se to get my final answer, But I did not resort to using a calculator either.. I noted that as x approaches two from both directions(left and right), cos(x-2) must approach 1 from values less than 1. Hence, 1-cos(x-2) must approach 0 from values greater than 0. So as the numerator is approaching 2, the denominator is approaching 0 from the right. This means the expression must approach infinity because values close to 2 are being divided by values that are infinitesimally close to and to the right of 0.
• Does that mean limit exist at x= 2?
• The limit does not exist at x=2 because the limit approaches infinity.

There a couple things you can notice right of the bat:
- The graph of cos(x) is continuous, while sec(x) is not.
- The slope of sec(x) is always the opposite sign of the slope of cos(x) (m ≠ 0).
- The y coordinates of cos(x) and sec(x) are never the same for the same x coordinate except when there is a relative max/min.
- Because the graphs of cos(x) and sec(x) always touch each other at their relative maximums and minimums, if you are given the graph of one you can visualize the graph of the other.

So while you can visualize the graph of cos(x) from looking at the graph of sec(x) (and vice versa), you cannot reliably do any calculations with sec(x) and expect to get the same result for cos(x).

I hope this helps!
• ¿How can the cosine be always positive?
• It isn't, but when x is close to 2 the value of `cos(x-2)` will be near `cos(0)` i.e. near to 1.

Most importantly, unless `x=2` the value of `cos(x-2)` will always be less than 1. This means that the denominator `1 - cos(x-2)` will always be positive for values of x that are close to, but not equal to 2.
• Does 0.99999999999999999999999999999999=1?
• No, it doesn't, unless you mean 0.9 repeating. 0.99999999999999999999999999999999 or whatever you said only approaches 1.
• Shouldn't the limit of the function, that is in the video, be either be undefined, unbounded, or nonexistent, instead of being equal to infinity? Also, isn't infinity not a number, but a concept. It is just so confusing. Can anyone help me understand?
• You're correct that the limit doesn't exist if the function goes to infinity. But if the function goes to infinity or negative infinity, we often like to write the limit as such because it provides more information. It's still implied that the limit doesn't exist, we're just giving info about the functions end behavior as well.
• so basically we are evaluating using linearization?
• How do you evaluate for a function like this:
f(x) = -x/ln^2(x-1)
I am trying to evaluate for when x approaches 2 from the left and right. I used the hints in the practice problems and did not understand how they obtained their values. Could someone please show me how to evaluate a function like this?

Thank You!
• Hi there
Use the idea that that ln(1) =0, and that for x>1, ln(x) is positive.

As x approaches 1 from the right, the values of ln(x) will become very small positive numbers.
So now, the numerator will have a value close to -1, while the denominator has a small positive value that you will square. The limit will be negative infinity.

Approach 1 from the left, and now ln(x) will approach a very small negative number - but when you square it, you will get a positive value - the left hand limit will equal the right hand limit. Now, while the limit "technically" doesn't exist at x = 1 because of the vertical asymptote, we do write that the limit is negative infinity... that at least tells us how the limit fails to exist and that the graph of the function has a vertical asymptote at x = 1.