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## Differential Calculus

### Course: Differential Calculus>Unit 1

Lesson 8: Strategy in finding limits

# Strategy 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.
A flow chart has options A through H, as follows. Step A, direct substitution. Try to evaluate the function directly. Evaluating f of a leads to options B through D. Option B: f of a = start fraction b divided by 0 end fraction, where b is not zero. The result is asymptote (probably). Example: the limit of start fraction 1 divided by x minus 1 end fraction as x approaches 1. Inspect with a graph or table to learn more about the function at x = a. Option C: f of a = b, where b is a real number. The result is limit found (probably). Example: limit of x squared as x approaches 3 = 3 squared = 9. Option D: f of a = start fraction 0 divided by 0 end fraction. Result is indeterminate form. Example: limit of start fraction x squared minus x minus 2 divided by x squared minus 2 x minus 3 end fraction, as x approaches negative 1. If you obtained option D, try rewriting the limit in an equivalent form. This leads to options E through G. Option E: factoring. Example: limit of start fraction x squared minus x minus 2 divided by x squared minus 2 x minus 3 end fraction, as x approaches negative 1 can be reduced to the limit of start fraction x minus 2 divided by x minus 3 end fraction as x approaches negative 1, by factoring and cancelling. Option F: conjugates. Example: the limit of start fraction start square root x end square root minus 2 divided by x minus 4 end fraction as x approaches 4 can be rewritten as the limit of start fraction 1 divided by start square root x end square root + 2 end fraction as x approaches 4, using conjugates and cancelling. Option G: trig identities. Example: limit of start fraction sine of x divided by sine of 2 x end fraction as x approaches 0 can be rewritten as the limit of start fraction 1 divided by 2 cosine of x end fraction as x approaches 0, using a trig identity. Using options E through G, try evaluating the limit in its new form, circling back to A, direct substitution. The last option is H, approximation: when all else fails, graphs and tables can help approximate 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

problem 1
g, left parenthesis, x, right parenthesis, equals, start fraction, x, minus, 3, divided by, square root of, x, plus, 5, end square root, minus, 3, end fraction
We want to find limit, start subscript, x, \to, 4, end subscript, g, left parenthesis, x, right parenthesis.
What happens when we use direct substitution?

Problem 2
h, left parenthesis, x, right parenthesis, equals, start fraction, 1, minus, cosine, left parenthesis, x, right parenthesis, divided by, 2, sine, squared, left parenthesis, x, right parenthesis, end fraction
We want to find limit, start subscript, x, \to, 0, end subscript, h, left parenthesis, x, right parenthesis.
What happens when we use direct substitution?

## Practice with the indeterminate form

Problem 3
Justin tried to find limit, start subscript, x, \to, minus, 1, end subscript, start fraction, x, plus, 1, divided by, x, squared, plus, 3, x, plus, 2, end fraction.
Using direct substitution, he got start fraction, 0, divided by, 0, end fraction.
For Justin's next step, which method would apply?

Problem 4
Catherine tried to find limit, start subscript, x, \to, minus, 3, end subscript, start fraction, square root of, 4, x, plus, 28, end square root, minus, 4, divided by, x, plus, 3, end fraction.
Using direct substitution, she got start fraction, 0, divided by, 0, end fraction.
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 limit, start subscript, x, \to, 5, end subscript, f, left parenthesis, x, right parenthesis for f, left parenthesis, x, right parenthesis, equals, start fraction, x, squared, minus, 25, divided by, x, squared, minus, 10, x, plus, 25, end fraction.
A flow chart has options A through H, as follows. Step A, direct substitution. Try to evaluate the function directly. Evaluating f of a leads to options B through D. Option B: f of a = start fraction b divided by 0 end fraction, here b is not zero. The result is asymptote (probably). Option C: f of a = b, where b is a real number. The result is limit found (probably). Option D: f of a = start fraction 0 divided by 0 end fraction. Result is indeterminate form. From option D, try rewriting the limit in an equivalent form. This leads to options E through G. Option E: factoring. Option F: conjugates. Option G: trig identities. Using options E through G, try evaluating the limit in its new form, circling back to A, direct substitution. The other option is H, approximation: when all else fails, graphs and tables can help approximate limits.
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 limit, start subscript, x, \to, 3, end subscript, f, left parenthesis, x, right parenthesis for f, left parenthesis, x, right parenthesis, equals, start fraction, square root of, 2, x, minus, 5, end square root, minus, 1, divided by, x, minus, 3, end fraction.
A flow chart has options A through H, as follows. Step A, direct substitution. Try to evaluate the function directly. Evaluating f of a leads to options B through D. Option B: f of a = start fraction b divided by 0 end fraction, here b is not zero. The result is asymptote (probably). Option C: f of a = b, where b is a real number. The result is limit found (probably). Option D: f of a = start fraction 0 divided by 0 end fraction. Result is indeterminate form. From option D, try rewriting the limit in an equivalent form. This leads to options E through G. Option E: factoring. Option F: conjugates. Option G: trig identities. Using options E through G, try evaluating the limit in its new form, circling back to A, direct substitution. The other option is H, approximation: when all else fails, graphs and tables can help approximate limits.
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.
• 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.
• I think it's important to add other indeterminate forms, such as ∞*0, as they appear sometimes in practices
• if you have limit as x goes to infinity as a question can you use any Real number?
• 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.
• what if its a e function like e^x or e^x-e? How would you try to solve it?