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. 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.

Worth being blunt about what that signpost does not say, because this is where beginners guess wrong:

0/0 is not a value that happens to be zero, and it is not a verdict that the limit fails to exist. It is the report you have learned nothing yet.

Three limits that all read 0/0 on substitution: x/x → 1, x²/x → 0, and x/x² → ∞ as x → 0⁺. Same form, three different answers. The form tells you which tool to reach for, never what the answer is.

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. A Taylor expansion rewrites a function near a point as a short polynomial that hugs it there. Replace each function by its first few series terms, such as:

  • 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.

The standard limits look narrow, but exams almost always hand you a scaled version. The move is always to force the pattern to appear.

Say you meet lim_{x → 0} sin(5x)/(2x). Write it as (5/2) · sin(5x)/(5x), so the second factor is exactly the standard limit with u = 5x and heads to 1. The answer is 5/2 = 2.5. Make the inside of the sin match the denominator, and pay for it with a constant out front.

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…
• 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? Could you turn to that rule 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 is not a value — it is a signpost to rewrite first, then take the limit:

  • 0/0
  • ∞ − ∞
  • ∞/∞

• 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