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### Course: Differential Calculus > Unit 1

Lesson 8: Strategy in finding limits# Strategy in finding limits

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

*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.(15 votes)
- Perfect! That's how you would do it, whereas putting in 0 in the first place gets you nowhere.(24 votes)

- This flow chart selection stuff is really frustrating. I know the answer exactly half the time I just can't put it in correctly.(15 votes)
- Yeah, the most frustrating quiz I have ever seen on Khan Academy.(8 votes)

- What if it's a ln(x)? Would it be conjugates, or trig identities? I think its conjugates, but how would you solve?(12 votes)
- You wouldn't use either, since those won't help you simplify the ln(), you would have to approximate.(4 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?(7 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.(7 votes)

- I'm wondering, why does the article say "probably"?

Is is it possible that f(x)=b/0 is not an asymptote, or that f(x) = b is not a real number? If so, what are some examples?(3 votes)- It could be the function is undefined for an interval.

It shouldn't be hard to come up with an example using piecewise functions.(8 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.(3 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)

- 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.(5 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)

- 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)