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Differential Calculus
Course: Differential Calculus > Unit 1
Lesson 8: Strategy in finding limitsStrategy in finding limits
AP.CALC:
LIM‑1 (EU)
, LIM‑1.E (LO)
, LIM‑1.E.1 (EK)
There are many techniques for finding limits that apply in various conditions. It's important to know all these techniques, but it's also important to know when to apply which technique.
Here's a handy dandy flow chart to help you calculate limits.
Key point #1: Direct substitution is the go-to method. Use other methods only when this fails, otherwise you're probably doing more work than you need to be. For example, it would be extra work to factor an expression into a simpler form if direct substitution would have worked without the factoring.
Key point #2: There's a big difference between getting b, slash, 0 and 0, slash, 0 (where b, does not equal, 0). When you get b, slash, 0, that indicates that the limit doesn't exist and is probably unbounded (an asymptote). In contrast, when you get 0, slash, 0, that indicates that you don't have enough information to determine whether or not the limit exists, which is why it's called the indeterminate form. If you wind up here, you've got more work to do, which is where the bottom half of the flow chart comes into play.
Note: There's a powerful method for finding limits called l'Hôpital's rule, which you'll learn later on. It's not covered here because we haven't learned about derivatives yet.
Practice with direct substitution
Practice with the indeterminate form
Putting it all together
Want to join the conversation?
For the problem 2, if factoring it by timing 1+cosx on numerator and denominator, I will get a new form 1/2(1+cosx). Then I can get limit 1/4. I am wondering if the result is correct.(8 votes)
Perfect! That's how you would do it, whereas putting in 0 in the first place gets you nowhere.(15 votes)
What if it's a ln(x)? Would it be conjugates, or trig identities? I think its conjugates, but how would you solve?(10 votes)
You wouldn't use either, since those won't help you simplify the ln(), you would have to approximate.(2 votes)
In problem 6, I think that there should be a step E (factoring) between step F (conjugates) and step A (direct substitution), because, in the numerator, 2x - 6 is factored into 2(x - 3) in order to cancel the (x - 3) term in the denominator. This may be insignificant but it seems correct to me. Am I right?(5 votes)
I'm with you, but I guess since the canceling out was done after the conjugates step they still count it as part of that step.(5 votes)
I think it's important to add other indeterminate forms, such as ∞*0, as they appear sometimes in practices(4 votes)- if you have limit as x goes to infinity as a question can you use any Real number?(2 votes)
- how to solve lim x tends to 5 sqrt(14-x) - 3 / sqrt(9-x) - 2(4 votes)
Nice problem!
I assume you mean
lim x tends to 5 of [sqrt(14-x) - 3]/[sqrt(9-x) - 2].
Direct substitution leads to the indeterminate form 0/0, so more work is required.
A good strategy is to multiply both top and bottom by the product of both the conjugate of the top and the conjugate of the bottom. This will create a pair of equal factors on top and bottom that cancel out.
lim x tends to 5 of [sqrt(14-x) - 3]/[sqrt(9-x) - 2].
= lim x tends to 5 of {[sqrt(14-x) - 3][sqrt(14-x) + 3][sqrt(9-x) + 2]}/{[sqrt(9-x) - 2][sqrt(14-x) + 3][sqrt(9-x) + 2]}
= lim x tends to 5 of {(14-x-9)[sqrt(9-x) + 2]}/{(9-x-4)[sqrt(14-x) + 3]}
= lim x tends to 5 of {(5-x)[sqrt(9-x) + 2]}/{(5-x)[sqrt(14-x) + 3]}
= lim x tends to 5 of [sqrt(9-x) + 2]/[sqrt(14-x) + 3]
= [sqrt(9-5) + 2]/[sqrt(14-5) + 3]
= (2+2)/(3+3)
= 2/3.(2 votes)
what if its a e function like e^x or e^x-e? How would you try to solve it?(3 votes)
When you have a limit of the type e^x, you would first have to substitute. No matter the value you plug in that function, it's going to be defined, so I don't see no problem.
With regard to e^(x-e), again the same. Substitute the value of the limit and you will find the desired solution.(3 votes)
If you find a real number with direct substitution why did you only probably find the limit?(2 votes)- You've only found the limit if the function is continuous at that point. If it isn't, the limit may be another value or may not exist.(4 votes)
- I'm asking a question which is not related to limits at all. I'm a high school student, and learning sequel, I wang to ask in terms of geometric sequel, can the sequel being 0,0,0,0,0,0,0,...? because the common ratio 0 is equal to 0/0 in this case, which is not defined.(3 votes)
So I came across a problem while studying for a test, and am stumped. what about the case where you have lim x-> 0 of sin(2x)/2x. I'm failing to figure out a rational way of turning this into a non indeterminate form.(2 votes)- Per SARAH684's comment, you can write the numerator as 2sin(x)cos(x), then cancel to get the limit of
[sin(x)/x] ·cos(x)
The limit as x goes to 0 of sin(x)/x is a special limit that requires more ingenuity to find. Sal finds it with a geometric diagram and the squeeze theorem here:
https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-8/v/sinx-over-x-as-x-approaches-0(2 votes)
For B and C in the flowchart, it amends Asymptote and Limit Found with the parenthetical (probably). Are there cases when this is not true, and direct substitution fails without producing an indeterminate form? Can I identify or check these cases without using a graph or table of the function?(2 votes)