- Topic
- Limits
- Difficulty
- Standard
- Suggested time
- 3 min
- Marks
- 2
Limits practice question 9
Evaluate the limit.
What do the top and the bottom each approach as ? With a square root in the way, how could you make a factor of appear on top?
Substitution gives . There is no factor to cancel yet, so create one.
Multiply the numerator and denominator by the conjugate .
A decimal to two decimal places or a fraction.
Theory UBC Math 100 §2.1.1CLP-1 §1.4Key results
Worked solution
Strategy
The indeterminate form comes from a difference of square roots. Multiplying by the conjugate turns that difference into a difference of squares, which cancels the troublesome factor.
Working
Substitution gives 0/0, so the expression has to be rewritten before the limit can be read off.
Multiply by the conjugate over itself. This is multiplication by 1, so the value is unchanged.
The numerator collapses by the difference-of-squares identity. Leave the denominator factored.
Cancel the shared factor of x. The indeterminate form is gone.
Substitute x = 0 into the continuous expression that remains.
Answer
The answer is 1/4, or 0.25.
Check
At x = 0.01 the quotient is (√4.01 − 2)/0.01 ≈ 0.2498, close to 1/4.
Where this goes wrong
Expanding the root term by term as √x + √4 and cancelling. The square root of a sum is not the sum of the square roots, so that step is not algebra you can do.
Leans on
Rationalising with a conjugate, from Functions, graphs and inverses
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