Limit Techniques

The previous lesson ended at a wall. You tried to evaluate lim_{x → 1} (x² − 1)/(x − 1) by substituting x = 1, and out came 0/0 — not an answer, just a shrug. Yet the curve plainly heads somewhere definite. Knowing what a limit means did not tell you how to compute one, and closing that gap is the whole of this lesson.

That 0/0 is an indeterminate form: the limit may well exist, but the expression has to be rewritten before it will show itself. ∞ − ∞ is another such form, and so is ∞/∞. None of them is a value — each is a signpost reading rewrite me first. This lesson is the small toolkit for that rewrite, and GATE DA leans on it hard: it recurs as a NAT, in 2024 and again in 2025. The same rewrites earn their keep in real machine-learning code, where a raw 0/0 or ∞ − ∞ becomes a NaN that silently poisons an entire training run.

Think of an indeterminate form as a locked door. 0/0 is not a dead end — it is a lock, and each lock takes a particular key. The craft is reading which lock you face, then reaching for the matching key rather than rattling all of them at once.

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 — exactly the wall from last lesson, now walked straight through.
• 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. Either way, the skill tested is the same: recognise the form, then choose 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. The form is a √-difference, so reach for the conjugate key: multiply by √(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 (pulling 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 feeding 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 promised.

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. This one is messy, so reach for the Taylor key: split the log of a product into a sum, then 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.

A question to carry forward

Each technique here works, but each also asks you to spot the trick — the right factor, the right conjugate, the right series. That cleverness is exactly what makes these problems feel slow under exam pressure. So a fair question arises: is there one mechanical rule that cracks any 0/0 or ∞/∞ without the hunt for an algebraic move — a rule you could turn almost without thinking? Here is the thread onward: what if, on an indeterminate quotient, you simply differentiated the top and the bottom separately and tried again?

In one breath

• An indeterminate form (0/0, ∞ − ∞, ∞/∞) is not a value — it is a signpost to rewrite first, then take the limit.
• Factor and cancel a 0/0 from a shared root: (x²−1)/(x−1) = x+1 → 2.
• Conjugate a √-difference / ∞ − ∞: √(t²+t) − t → 1/2 (GATE DA 2025).
• Standard limits, quotable on sight: sin x / x → 1 and (eˣ − 1)/x → 1 as x → 0.
• Taylor-expand messy forms (sin x ≈ x, cos x ≈ 1 − x²/2, ln(1+u) ≈ u): the 2024 ln((x²+1)cos x)/x² → 1/2 falls in two lines.
• Reflex: read the form, then pick the matching key — never treat ∞ − ∞ as plain subtraction.

Practice