Limits and Derivatives · Lesson 4 of 8
Trigonometric Limits and the Squeeze Idea
“Use geometric bounds to establish the limits that later make trigonometric differentiation possible.”
• Explain how inequalities between functions constrain their limits. • Apply the Sandwich Theorem. • Follow the triangle–sector comparison proving the standard sine limit. • Derive the related cosine and tangent limits. • Rescale and shift angles to evaluate trigonometric limits in radians.
Bounding a function instead of evaluating it directly
Sometimes a difficult function can be trapped between two simpler ones. If the bounds approach the same number, the trapped function has nowhere else to approach. This idea is particularly useful for sin x/x, where direct substitution at zero gives 0/0 and no algebraic polynomial factorisation is available.
First consider two functions f and g with f(x)≤g(x) for all relevant inputs sufficiently near a. If both finite limits exist, the limiting values preserve that order. They can become equal even if the nearby values are different, so a strict inequality need not remain strict in the limit.
Now place g between f and h. If the lower and upper functions both approach L, then g must approach L too. This is called the Sandwich Theorem, also commonly called the Squeeze Theorem. It establishes the middle limit rather than assuming in advance that it exists.
Problem
Suppose 2−x²≤g(x)≤2+x² for all non-zero x close to zero. Find the limit of g(x).
- 1.The lower polynomial has limit 2−0=2.
- 2.The upper polynomial has limit 2+0=2.
- 3.Since g stays between bounds with the same limit, the Sandwich Theorem gives limit 2. A separate formula for g is unnecessary.
The geometry behind the sine limit
Take a unit circle with centre O. Put A at the right end of a horizontal radius and C on the circle at an angle x above OA, where 0<x<π/2. Draw the tangent through A and extend OC until it meets that tangent at B. Drop a perpendicular from C to OA at D.
The small triangle OAC lies inside the sector OAC, and that sector lies inside the larger triangle OAB. Comparing their areas will compare sin x, the angle x, and tan x. The sector-area formula is especially simple because x is measured in radians.
With unit radius, the small triangle has base 1 and height sin x, so its area is sin x/2. The sector has area x/2, because a sector of radius r and angle x radians has area r²x/2. The large triangle has base 1 and height tan x, so its area is tan x/2. Multiply the area inequalities by 2 to get sin x<x<tan x.
All three quantities are positive for this range of angles. Dividing by sin x gives 1<x/sin x<1/cos x. Taking reciprocals reverses the inequalities, giving the bounds below.
The ratio sin x/x is unchanged when x is replaced by −x, because both numerator and denominator change sign. Cosine is also unchanged. Therefore the same bounds hold for negative x close to zero, with 0<|x|<π/2. Both bounding functions approach 1 as x approaches zero from either side. The Sandwich Theorem now proves the standard sine limit.
The standard sine limit has value 1 when the angle input is in radians. Do not substitute degree measurements into the same formula. Also, sin x/x is not defined at zero; its limit is determined by nearby non-zero angles.
Related limits from the standard result
Use the identity 1−cos x=2sin²(x/2). This turns the cosine expression into one standard sine ratio multiplied by a sine term. As x approaches zero, x/2 does too; the ratio approaches 1 and the remaining sine approaches zero.
Tangent equals sin x/cos x wherever cosine is non-zero. Near zero, cos x approaches 1, so tan x/x is the product of sin x/x and 1/cos x. These limits give 1×1=1. The result follows from an identity and the quotient law rather than a separate geometric proof.
Problem
Evaluate lim[x→0] tan x/x.
- 1.Write tan x/x=(sin x/x)(1/cos x).
- 2.The first factor tends to 1 by the standard sine limit.
- 3.The second tends to 1 because cos x tends to 1, a non-zero number.
- 4.The product has limit 1.
Match the angle to the denominator
A sine ratio becomes standard only when its denominator matches its angle. For sin(4x), the matching denominator is 4x rather than x. Insert the missing constant deliberately, then track it outside the limit. This prevents a common error in scaled-angle problems.
Problem
Evaluate lim[x→0] sin(4x)/sin(2x).
- 1.For nearby non-zero x, write the ratio as 2[sin(4x)/(4x)]/[sin(2x)/(2x)].
- 2.As x→0, both 4x and 2x approach zero. Each matched sine ratio approaches 1.
- 3.The result is 2×1/1=2.
Problem
Find lim[x→0] sin(ax)/sin(bx) for non-zero constants a and b.
- 1.Write the expression as (a/b)[sin(ax)/(ax)]/[sin(bx)/(bx)].
- 2.Both matched ratios approach 1, including when a or b is negative, because the standard limit is two-sided.
- 3.The result is a/b. The constants do not disappear from the calculation.
Problem
Evaluate lim[x→π/2] tan(2x)/(x-π/2).
- 1.Set u=x−π/2, so u→0 and 2x=π+2u.
- 2.The periodicity identity gives tan(π+2u)=tan(2u). The quotient becomes 2[tan(2u)/(2u)].
- 3.The tangent ratio tends to 1, so the required limit is 2. Although tan x is undefined at π/2, the numerator here is tan(2x), a different function.
When a trigonometric quotient gives 0/0, identities may reveal the useful cancellation. For example, cos(2x)−1=−2sin²x and cos x−1=−2sin²(x/2). Their ratio is a squared scaled-sine ratio and therefore approaches 4. Always use identities before assuming that two zero expressions cancel.
Quiz
What must the two outside functions in the Sandwich Theorem do?
What is the limit of sin x/x at zero in radians?
What is the limit of (1−cos x)/x at zero?
What is the limit of sin(6x)/sin(3x) at zero?
Why does the sine proof work on both sides of zero?
What is the limit of tan x/x at zero?
Practice Problems
- Evaluate lim[x→0] sin(ax)/(bx) for non-zero b. Discuss also the case a=0.
- Evaluate lim[x→0] sin(ax)/sin(bx) for a,b non-zero.
- Evaluate lim[x→π] sin(π-x)/[π(π-x)] by introducing u=π−x.
- Evaluate lim[x→0] cos x/(π-x).
- Evaluate lim[x→0] [cos(2x)-1]/[cos x-1] using half-angle identities.
- Evaluate lim[x→0] [ax+x·cos x]/[b·sin x] for non-zero b.
- Evaluate lim[x→0] x·sec x.
- Evaluate lim[x→0] [sin(ax)+bx]/[ax+sin(bx)] when a,b and a+b are non-zero.
- Evaluate lim[x→0] (cosec x-cot x). First combine the terms as one fraction.
- Evaluate lim[x→π/2] tan(2x)/(x-π/2), explaining the shift of angle.
Key Takeaways
• Order is preserved by existing finite limits, although a strict inequality can become equality. • Matching lower and upper limits force the middle function to share that limit. • Unit-circle area comparisons establish the standard sine limit in radians. • The cosine and tangent limits follow from identities and the sine limit. • For scaled or shifted angles, rewrite the expression to match a standard limit before substituting.