Standard Trigonometric Limits
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Direct answer
Three limits — sin x/x → 1, (1 − cos x)/x² → 1/2 and tan x/x → 1 as x → 0 — power nearly every trigonometric limit in JEE, because every other one is assembled from them by algebra, conjugates or series. The squeeze theorem proves the first geometrically on the unit circle, and that proof is why sin x/x → 1 holds only in radians. The offshoots: sin(ax)/x → a, (1 − cos ax)/x² → a²/2, (sin ax)/(sin bx) → a/b, and (1 − cos x)/x → 0, not 1/2. Non-trig companions share the toolbox: (e^x − 1)/x → 1, ln(1 + x)/x → 1 and (1 + f)^(1/f) → e whenever f → 0.
What you must remember
- The trio: sin x/x → 1, tan x/x → 1, (1 − cos x)/x² → 1/2 as x → 0, all in radians.
- Scaled forms: sin 5x/x → 5; (1 − cos 3x)/x² → 9/2; (tan 2x)/(sin 5x) → 2/5 — match each argument to its own denominator factor.
- Wrong-denominator trap: (1 − cos x)/x → 0, since 1 − cos x ~ x²/2 is one order smaller than x.
- Companion exponential and log limits: (e^x − 1)/x → 1, (a^x − 1)/x → ln a, ln(1 + x)/x → 1.
- Squeeze theorem: if g ≤ f ≤ h near the point and g, h share a limit L, then f → L — the machine behind sin x/x and behind lim x→0 x² sin(1/x) = 0, which has no other route.
- 1^∞ indeterminate: bases tending to 1 with exponents blowing up resolve as e^(limit of exponent × ln(base)).
- Half-angle bridge: 1 − cos x = 2 sin²(x/2) converts any (1 − cos)/power limit into pure sine form.
When the standard three are not enough
Evaluate lim x→0 (tan x − sin x)/x³. Factor before expanding: tan x − sin x = tan x(1 − cos x), so the expression splits as (tan x/x) × ((1 − cos x)/x²) = 1 × 1/2 = 1/2. The standard trio does the whole job once the algebra exposes them — but suppose the denominator had been x² instead: then tan x − sin x ≈ x³/2 divided by x² gives 0, and recognising the order gap replaces computation. For the harder tier, series carry further than the trio: tan x = x + x³/3 + ... and sin x = x − x³/6 + ..., so tan x − sin x = x³/2 + higher terms — same answer, and this route survives when factoring fails, as in lim (x − sin x)/x³ = 1/6, where no algebraic regrouping produces the standard forms at all. The professional habit: try the trio by factoring first, and the moment two rounds of rearrangement have not exposed them, switch to expansions rather than looping.
Examiner's angle
JEE Main asks the plug-in-and-rearrange type: (sin 3x)/(tan 5x) → 3/5, (1 − cos 2x)/(x tan x) → 2, and the 1^∞ classics like (cos x)^(1/x²) → e^(−1/2), since ln(cos x) ≈ −x²/2. The distractors are the twin values 1/2 versus 1/4 from (1 − cos kx)/x² with the argument doubled, and 0 versus the genuine value for (1 − cos x)/x. JEE Advanced builds limits needing the expansion tier — (tan x − sin x)/x³, (x − sin x)/x³, cos(sinx)... and the exam's historical fondness for e^(−1/6) from (sin x/x)^(1/x²) makes that value worth memorising as a final answer, not just a method. Squeeze-theorem items hide as products like x² cos(1/x²), where oscillation kills every algebraic approach and only bounding does the work. Guard the radians condition: a limit stated in degrees is a deliberate misdirection, since sin x°/x then tends π/180, and one option always carries the innocent-looking 1.
Frequently asked questions
Why does sin x/x → 1 only in radians?
The squeeze proof compares sin x with the arc length x itself; arcs are measured in radians, so the geometric sandwich breaks for degrees.
What is lim (1 − cos x)/x² as x → 0?
1/2, via 1 − cos x = 2 sin²(x/2); dividing by x instead gives 0.
How do scaled arguments behave?
sin(ax)/x → a and (1 − cos ax)/x² → a²/2 — each argument's factor survives division by its matching power.
When is the squeeze theorem the only route?
When an oscillating bounded factor multiplies something tending to 0, as in x² sin(1/x) or x cos(1/x²), where algebra and series both fail.
How is a 1^∞ form resolved?
Rewrite as e^(g(x)·ln(f(x))) and evaluate the exponent's limit, standard practice for items like (cos x)^(1/x²).