EAPCET Mathematics · Intermediate first year · Paper IB

Limits and Continuity

Limits and Continuity introduces the limit of a function and the conditions for a function to be continuous — the entry point to calculus.

Subject
Mathematics
Taught in
Intermediate first year · Paper IB
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • lim⁡x→0sin⁡xx=1\lim_{x \to 0} \dfrac{\sin x}{x} = 1
  • lim⁡x→0ex−1x=1\lim_{x \to 0} \dfrac{e^{x} - 1}{x} = 1 and lim⁡x→0ax−1x=log⁡a\lim_{x \to 0} \dfrac{a^{x} - 1}{x} = \log a
  • lim⁡x→0(1+x)1/x=e\lim_{x \to 0} (1 + x)^{1/x} = e
  • lim⁡x→axn−anx−a=nan−1\lim_{x \to a} \dfrac{x^{n} - a^{n}}{x - a} = na^{n-1}
  • ff is continuous at aa when the left limit, right limit and f(a)f(a) are all equal
  • 1∞1^{\infty} form: if f(x)→0f(x) \to 0 and g(x)→∞g(x) \to \infty, then lim⁡(1+f(x))g(x)=elim⁡f(x)g(x)\lim (1 + f(x))^{g(x)} = e^{\lim f(x)g(x)}
  • As x→0x \to 0: lim⁡sin⁡pxx=p\lim \dfrac{\sin px}{x} = p, lim⁡tan⁡kxx=k\lim \dfrac{\tan kx}{x} = k, lim⁡log⁡(1+px)x=p\lim \dfrac{\log(1 + px)}{x} = p
  • As x→0x \to 0: lim⁡1−cos⁡xx2=12\lim \dfrac{1 - \cos x}{x^{2}} = \dfrac{1}{2}
  • sin⁡x=x−x33!+x55!−…\sin x = x - \dfrac{x^{3}}{3!} + \dfrac{x^{5}}{5!} - \dots; cos⁡x=1−x22!+x44!−…\cos x = 1 - \dfrac{x^{2}}{2!} + \dfrac{x^{4}}{4!} - \dots; tan⁡x=x+x33+2x515+…\tan x = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \dots
  • sin⁡−1x=x+x36+3x540+…\sin^{-1}x = x + \dfrac{x^{3}}{6} + \dfrac{3x^{5}}{40} + \dots; tan⁡−1x=x−x33+x55−…\tan^{-1}x = x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - \dots
  • ex=1+x+x22!+x33!+…e^{x} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + \dots; ax=1+xlog⁡a+(xlog⁡a)22!+…a^{x} = 1 + x\log a + \dfrac{(x\log a)^{2}}{2!} + \dots
  • log⁡(1+x)=x−x22+x33−…\log(1 + x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \dots; log⁡(1−x)=−(x+x22+x33+… )\log(1 - x) = -\left(x + \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} + \dots\right)
  • (1+x)n=1+nx+n(n−1)x22!+…(1 + x)^{n} = 1 + nx + \dfrac{n(n - 1)x^{2}}{2!} + \dots for ∣x∣<1|x| < 1

What this chapter covers

  • Left and right limits
  • Standard limits
  • Limits at infinity
  • Continuity at a point and on an interval

How it is usually asked

  • Evaluate a limit of indeterminate form
  • Find the value of a constant that makes a function continuous
  • Evaluate a limit at infinity

Understanding Limits and Continuity

The limit of f(x) as x approaches a is the value f(x) gets close to when x is near a, regardless of what happens exactly at a. The limit exists only when the left-hand limit and the right-hand limit both exist and are equal. This is why functions involving |x|, the greatest integer function or piecewise definitions need the two sides checked separately.

Most EAPCET limits are of the indeterminate form 0/0 or ∞/∞, and the job is to rewrite them until a standard limit appears. Factorise and cancel, rationalise expressions with square roots, or reshape the expression into one of the standard forms: sin x/x → 1, (eˣ − 1)/x → 1, (aˣ − 1)/x → log a and (1 + x)^(1/x) → e as x → 0. For limits at infinity, divide by the highest power of x.

A function is continuous at a when the limit at a exists and equals f(a). In practice this means the left limit, the right limit and the function value all match. A typical question defines a function piecewise with an unknown constant and asks for the value that makes it continuous; you compute the limit and set the constant equal to it.

Worked example

EAPCET-style question

f(x)=1−cos⁡4xx2f(x) = \dfrac{1 - \cos 4x}{x^{2}} for x≠0x \ne 0 and f(0)=kf(0) = k. If ff is continuous at x=0x = 0, then kk equals

  1. (1)2
  2. (2)4
  3. (3)8
  4. (4)16

Solution

  1. Continuity needs k=lim⁡x→01−cos⁡4xx2k = \lim_{x \to 0} \dfrac{1 - \cos 4x}{x^{2}}.
  2. 1−cos⁡4x=2sin⁡22x1 - \cos 4x = 2\sin^{2}2x, so the expression is 2(sin⁡2xx)22\left(\dfrac{\sin 2x}{x}\right)^{2}.
  3. sin⁡2xx=2⋅sin⁡2x2x→2\dfrac{\sin 2x}{x} = 2 \cdot \dfrac{\sin 2x}{2x} \to 2, so the limit is 2×22=82 \times 2^{2} = 8.
  4. Therefore k=8k = 8.

Answer: 8

Shortcuts and results to remember

  • For 00\tfrac{0}{0} or ∞∞\tfrac{\infty}{\infty} use L’Hospital’s rule or series expansions; for ∞−∞\infty - \infty take an LCM or rationalise to convert it to 00\tfrac{0}{0}.
  • Rationalise when the expression has square roots, and factorise when it is a ratio of polynomials.
  • In trig limits use 1+cos⁡2x=2cos⁡2x1 + \cos 2x = 2\cos^{2}x and 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^{2}x to remove the 00\tfrac{0}{0}.
  • Polynomials are continuous everywhere; for functions with [x][x] or ∣x∣|x|, compare the left and right limits at the break points.

More practice questions

Practice question 1

The value of lim⁡x→0(1+3x)2/x\lim_{x \to 0} (1 + 3x)^{2/x} is

  1. (1)e6e^{6}
  2. (2)e3e^{3}
  3. (3)e2/3e^{2/3}
  4. (4)1

Solution

  1. This is the 1∞1^{\infty} form with f(x)=3xf(x) = 3x and g(x)=2xg(x) = \dfrac{2}{x}
  2. Limit =elim⁡f(x)g(x)= e^{\lim f(x)g(x)}
  3. f(x) g(x)=3x×2x=6f(x)\,g(x) = 3x \times \dfrac{2}{x} = 6
  4. Limit =e6= e^{6}

Answer: e6e^{6}

Practice question 2

The value of lim⁡x→0x−sin⁡xx3\lim_{x \to 0} \dfrac{x - \sin x}{x^{3}} is

  1. (1)16\dfrac{1}{6}
  2. (2)13\dfrac{1}{3}
  3. (3)0
  4. (4)12\dfrac{1}{2}

Solution

  1. sin⁡x=x−x36+x5120−…\sin x = x - \dfrac{x^{3}}{6} + \dfrac{x^{5}}{120} - \dots
  2. x−sin⁡x=x36−x5120+…x - \sin x = \dfrac{x^{3}}{6} - \dfrac{x^{5}}{120} + \dots
  3. Divide by x3x^{3}: 16−x2120+…\dfrac{1}{6} - \dfrac{x^{2}}{120} + \dots
  4. As x→0x \to 0 the limit is 16\tfrac{1}{6}

Answer: 16\dfrac{1}{6}

Common mistakes to avoid

  • Using sin x/x → 1 with x in degrees; the standard limit holds only for radian measure.
  • Computing one side of a limit at a point where the function changes formula, such as for [x] or |x|; check both one-sided limits.
  • Treating the 1^∞ form as equal to 1; limits like (1 + 2x)^(1/x) give e², not 1.
  • For limits at infinity, dividing by a power of x lower than the highest one in the denominator, which leaves an ∞/∞ form.

How to prepare it

Reduce every limit to a standard form — factorise, rationalise or substitute before reaching for anything else.

For continuity questions, write the left limit, right limit and function value separately and equate them.

Once the formulas feel familiar, test them the way the exam does: practise Limits and Continuity questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Limits and Continuity chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Limits and Continuity part of the EAPCET Mathematics syllabus?
Yes. Limits and Continuity is taught in Intermediate first year · Paper IB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Limits and Continuity for EAPCET?
Reduce every limit to a standard form — factorise, rationalise or substitute before reaching for anything else. For continuity questions, write the left limit, right limit and function value separately and equate them.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Limits and Continuity?
Using sin x/x → 1 with x in degrees; the standard limit holds only for radian measure. Computing one side of a limit at a point where the function changes formula, such as for [x] or |x|; check both one-sided limits. Treating the 1^∞ form as equal to 1; limits like (1 + 2x)^(1/x) give e², not 1. For limits at infinity, dividing by a power of x lower than the highest one in the denominator, which leaves an ∞/∞ form.