A student wants to find lim(x→0) x²·sin(1/x). They note that sin(1/x) oscillates wildly near 0 and conclude the limit does not exist. Which approach correctly finds the limit?
ADirect substitution: 0²·sin(1/0) = 0, so the limit is 0
BSince −x² ≤ x²sin(1/x) ≤ x² and both ±x² → 0 as x → 0, the Squeeze Theorem gives the limit as 0
CApply L'Hôpital's rule to the expression to resolve the oscillation
DThe limit does not exist because sin(1/x) has no limit as x → 0
Even though sin(1/x) oscillates infinitely near 0, we always have −1 ≤ sin(1/x) ≤ 1, which gives −x² ≤ x²sin(1/x) ≤ x². Both bounding functions approach 0, so the Squeeze Theorem forces the limit to 0. The oscillation of sin(1/x) is irrelevant — the bounding functions trap f regardless. L'Hôpital's rule doesn't apply here (x²sin(1/x) is not in 0/0 or ∞/∞ form), and option D ignores that f can still have a limit even when a factor does not.
Question 2 Multiple Choice
To apply the Squeeze Theorem to find lim(x→a) f(x), a student finds g(x) ≤ f(x) ≤ h(x) near x = a, with lim g(x) = 2 and lim h(x) = 3. What conclusion is valid?
Alim f(x) = 2.5, the midpoint of the two limiting values
Blim f(x) = 2, since g is the lower bound
CThe Squeeze Theorem cannot be applied; the bounding limits must be equal
D2 ≤ lim f(x) ≤ 3, a valid squeeze-theorem conclusion
The Squeeze Theorem requires that both bounding functions converge to the *same* limit L. When lim g(x) = 2 ≠ 3 = lim h(x), the theorem gives no conclusion — f could approach any value in [2, 3], or the bounds might not trap f tightly enough to determine the limit at all. Option D sounds reasonable but is NOT what the Squeeze Theorem provides; it only applies when both limits are equal. The theorem is an equality result, not an interval result.
Question 3 True / False
The Squeeze Theorem can be used to find lim(x→0) sin(x)/x = 1 even though sin(x)/x is undefined at x = 0.
TTrue
FFalse
Answer: True
The Squeeze Theorem only requires the bounding inequality g(x) ≤ f(x) ≤ h(x) to hold *near* x = a — not necessarily at x = a itself. The function sin(x)/x is undefined at 0, but the bounds cos(x) ≤ sin(x)/x ≤ 1 hold for all small nonzero x in radians. Since both bounds converge to 1 as x → 0, the limit is 1. This is precisely the power of limit arguments: behavior at a point is irrelevant to the limit.
Question 4 True / False
If g(x) ≤ f(x) ≤ h(x) near x = a, and lim(x→a) g(x) = lim(x→a) h(x) = L, then f(x) = L for most x near a.
TTrue
FFalse
Answer: False
The Squeeze Theorem concludes that lim(x→a) f(x) = L — a statement about the *limit*, not about the function's values. f(x) can oscillate, dip, or spike away from L at individual points near a, as long as it is trapped between g and h and those bounds squeeze toward L. For example, x²sin(1/x) oscillates wildly near 0 but has limit 0. Confusing a limit with a function value is a fundamental error.
Question 5 Short Answer
Why does proving lim(x→0) sin(x)/x = 1 require geometric reasoning rather than algebraic manipulation?
Think about your answer, then reveal below.
Model answer: Direct substitution gives 0/0 — an indeterminate form that algebraic simplification cannot resolve, because sin(x)/x has no simpler closed form at x = 0. The geometric argument uses the unit circle to establish the inequality cos(x) ≤ sin(x)/x ≤ 1 for small positive x in radians, derived by comparing the areas of a triangle, a circular sector, and a larger triangle. The Squeeze Theorem then takes over from there, forcing the limit to 1 since both bounds converge to 1.
This limit is foundational because it underlies the derivative of sin(x). The geometric proof reveals *why* the limit is 1 — it is a consequence of how arc length compares to chord length on the unit circle, not an algebraic accident. The bounding strategy (find a function always ≥ f and always ≤ f with the same limit) is the core skill: once you have the bounds, the Squeeze Theorem is automatic.