The Limit Laws and Indeterminate Forms · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Taming zero over zero

Mathematics · Calculus & Analysis · ages 18-19
Name ______________________   Date ____________

  1. Substitution into a quotient gives 0/0, so its limit does not exist.

    Circle one:   True   False

  2. The only information about g near 0 is -1 ≤ g(x) ≤ 1. What does this establish about its limit?

    • No unique limit is established.
    • Its limit must equal the lower bound -1.
    • Its limit must equal the upper bound 1.
    • Its limit must equal the mean of the bounds.

  3. When simplifying (x-4)(x+1)/(x-4), under which input condition is cancelling x-4 valid?

    • Only when x is exactly 4.
    • Only when x is greater than 4.
    • Whenever x is distinct from 4.
    • Only when x+1 is also zero.

  4. Suppose f tends to 5 and g tends to -1. What does fg+3 tend to?

    • -8
    • -2
    • 2

  5. Using radians, find the limit of sin(4x)/x as x approaches 0.

    Answer: ______________

  6. Find the limit of x/(sqrt(1+x)-1) as x approaches 0.

    Answer: ______________

  7. What is the limit of (sqrt(x)-2)/(x-4) as x approaches 4? Give an exact decimal.

    Answer: ______________

  8. Find the limit of 4+x sin(1/x) as x approaches 0.

    Answer: ______________

  9. Using radians, find the limit of tan(2x)/x as x approaches 0.

    Answer: ______________

  10. What is the limit of x²sin(1/x) as x approaches 0, and why?

    • It is absent because multiplying cannot reduce the sine oscillation.
    • It is 0 because its absolute value is at most x².
    • It is 1 because every sine quotient has limit 1.
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Answer key

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Taming zero over zero W1-mt_rAJs-V4Jlx-s1

  1. False · The form is indeterminate; x/x at zero gives 0/0 by substitution but has limit 1.
  2. No unique limit is established. · The bounds have different limits; constant functions 0 and 1 both satisfy them.
  3. Whenever x is distinct from 4. · Cancellation divides by x-4, so that factor must be nonzero.
  4. -2 · Multiply the component limits, then add 3: -5+3 = -2.
  5. 4 · Rewrite as 4 sin(4x)/(4x); the quotient tends to 1.
  6. 2 · Rationalising the denominator leaves sqrt(1+x)+1, with limit 2.
  7. 0.25 · The conjugate leaves 1/(sqrt(x)+2), with limit 1/4 = 0.25.
  8. 4 · The added oscillating term has magnitude at most |x|, so it tends to zero.
  9. 2 · The expression is [sin(2x)/(2x)][2/cos(2x)], whose factors tend to 1 and 2.
  10. It is 0 because its absolute value is at most x². · The output magnitude is bounded by x², which tends to zero.
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