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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.
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 $b/0$ and $0/0$ (where $b\ne 0$). When you get $b/0$, that indicates that the limit doesn't exist and is probably unbounded (an asymptote). In contrast, when you get $0/0$, 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

problem 1
$g\left(x\right)=\frac{x-3}{\sqrt{x+5}-3}$
We want to find $\underset{x\to 4}{lim}g\left(x\right)$.
What happens when we use direct substitution?

Problem 2
$h\left(x\right)=\frac{1-\mathrm{cos}\left(x\right)}{2{\mathrm{sin}}^{2}\left(x\right)}$
We want to find $\underset{x\to 0}{lim}h\left(x\right)$.
What happens when we use direct substitution?

Practice with the indeterminate form

Problem 3
Justin tried to find $\underset{x\to -1}{lim}\frac{x+1}{{x}^{2}+3x+2}$.
Using direct substitution, he got $\frac{0}{0}$.
For Justin's next step, which method would apply?

Problem 4
Catherine tried to find $\underset{x\to -3}{lim}\frac{\sqrt{4x+28}-4}{x+3}$.
Using direct substitution, she got $\frac{0}{0}$.
For Catherine's next step, which method would apply?

Putting it all together

Problem 5
Jill's teacher gave her a flow chart (below) and asked her to find $\underset{x\to 5}{lim}f\left(x\right)$ for $f\left(x\right)=\frac{{x}^{2}-25}{{x}^{2}-10x+25}$.
Drag the cards below to show Jill's path to finding the limit.
A. Direct substitution
B. Asymptote
C. Limit found
D. Indeterminate form
E. Factoring
F. Conjugates
G. Trig identities
H. Approximation

Problem 6
Fenyang's teacher gave him a flow chart (below) and asked him to find $\underset{x\to 3}{lim}f\left(x\right)$ for $f\left(x\right)=\frac{\sqrt{2x-5}-1}{x-3}$.
Drag the cards below to show Fenyang's path to finding the limit.
A. Direct substitution
B. Asymptote
C. Limit found
D. Indeterminate form
E. Factoring
F. Conjugates
G. Trig identities
H. Approximation

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.
• Perfect! That's how you would do it, whereas putting in 0 in the first place gets you nowhere.
• What if it's a ln(x)? Would it be conjugates, or trig identities? I think its conjugates, but how would you solve?
• You wouldn't use either, since those won't help you simplify the ln(), you would have to approximate.
• This flow chart selection stuff is really frustrating. I know the answer exactly half the time I just can't put it in correctly.
• Yeah, the most frustrating quiz I have ever seen on Khan Academy.
• 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?
• It could be the function is undefined for an interval.

It shouldn't be hard to come up with example using piecewise functions
• 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?
• 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.
• how to solve lim x tends to 5 sqrt(14-x) - 3 / sqrt(9-x) - 2
• 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.
• I think it's important to add other indeterminate forms, such as ∞*0, as they appear sometimes in practices