Limit Techniques

You know what a limit means. Now you have to actually compute one — and the moment you try, you hit a wall. Plugging in the target gives 0/0 or ∞ − ∞, which says nothing.

Those are indeterminate forms: the limit may still exist, but you have to rewrite the expression first to see it. This lesson is the small toolkit for that rewrite. GATE DA leans on it hard — it shows up as a recurring NAT, in 2024 and again in 2025. The same rewrites keep you out of trouble in real ML code, where a raw 0/0 or ∞ − ∞ becomes a NaN that silently poisons a whole training run.

The four techniques

• Factor and cancel. When direct substitution gives 0/0 because numerator and denominator share a root, factor both and cancel the common piece. The classic (x² − 1)/(x − 1) becomes (x − 1)(x + 1)/(x − 1) = x + 1 → 2 as x → 1.
• Rationalise with the conjugate. When a √ difference produces 0/0 or ∞ − ∞, multiply top and bottom by the conjugate (same terms, opposite sign) to turn the square-root difference into a clean difference of squares.
• Standard limits. Memorise these two — they appear constantly: sin x / x → 1 as x → 0, and (eˣ − 1)/x → 1 as x → 0.
• Taylor expansion. Replace each function by its first few series terms near the point (e.g. sin x ≈ x, cos x ≈ 1 − x²/2, ln(1 + u) ≈ u − u²/2, eˣ ≈ 1 + x), then cancel and read off the ratio. This dissolves messy 0/0 forms quickly.

How GATE asks this

This is a NAT favourite: you are handed a limit that substitution cannot crack and must type the numeric value. GATE DA 2025 asked lim_{t → ∞} (√(t² + t) − t) — a ∞ − ∞ form cracked by the conjugate (worked below). GATE DA 2024 asked lim_{x → 0} ln((x² + 1)·cos x) / x² — a 0/0 form cracked by Taylor expansion. Occasionally the form appears as an MCQ asking which technique applies. The skill being tested is recognising the form and choosing the right rewrite.

Worked example 1 — a real GATE DA 2025 question

Evaluate lim_{t → ∞} (√(t² + t) − t).

Substituting t = ∞ gives ∞ − ∞, which is indeterminate — do not stop here. Multiply by the conjugate √(t² + t) + t over itself:

√(t²+t) − t  =  (√(t²+t) − t) · (√(t²+t) + t)
                ──────────────────────────────
                        √(t²+t) + t

             =      (t² + t) − t²              t
                ──────────────────────  =  ───────────
                     √(t²+t) + t            √(t²+t) + t

Now divide top and bottom by t (and pull t out of the root as √(t²·…) = t·√…):

        t                       1
   ───────────────  =  ────────────────────   →   1 / (√1 + 1)  =  1/2
   √(t²+t) + t            √(1 + 1/t) + 1

As t → ∞, the 1/t inside the root vanishes, leaving 1/(√1 + 1) = 1/2. So the limit is 0.5. (This is GATE DA 2025.)

Sanity-check it by plugging in growing values of t: at t = 10 the expression is already 0.488…, at t = 100 it is 0.498…, and at t = 10⁵ it is 0.499999. The march toward 0.5 is exactly what the algebra predicted.

Worked example 2 — a real GATE DA 2024 question

Evaluate lim_{x → 0} ln((x² + 1)·cos x) / x².

Substituting x = 0 gives ln(1·1)/0 = 0/0 — indeterminate. Split the log of a product into a sum, then Taylor-expand each piece near x = 0:

ln((x²+1)·cos x) = ln(1 + x²) + ln(cos x)

   ln(1 + x²) ≈ x²                 (since ln(1+u) ≈ u, with u = x²)
   ln(cos x)  ≈ ln(1 − x²/2) ≈ −x²/2   (since cos x ≈ 1 − x²/2)

Add the two expansions and divide by x²:

ln((x²+1)·cos x)      x² − x²/2        x²/2      1
─────────────────  ≈  ───────────  =  ──────  =  ──  =  0.5
       x²                 x²            x²        2

The higher-order terms vanish faster than x², so the limit is exactly 0.5. (This is GATE DA 2024.) Two functions, two short series, and the 0/0 dissolves.

Quick check