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Concept 01 / 12Limits, Continuity and Differentiability

Limits: Meaning, One-Sided Limits and Existence

The intuitive and epsilon-delta meaning of a limit, why the value at the point is irrelevant, left-hand and right-hand limits, when a limit exists, limits at infinity and infinite limits, and evaluating one-sided limits of piecewise and greatest-integer functions.

Figure 1
The statement lim⁡x→af(x)=L\lim_{x \to a}f(x) = L means that f(x)f(x) can be made as close to LL as we like by taking xx close enough to aa, but not equal to aa. The precise version: for every ε>0\varepsilon > 0 there is a δ>0\delta > 0 such that 0<∣x−a∣<δ0 < |x - a| < \delta implies ∣f(x)−L∣<ε|f(x) - L| < \varepsilon. Two features of this definition matter constantly. First, the value f(a)f(a) plays no role. The condition 0<∣x−a∣0 < |x - a| excludes x=ax = a itself. So ff need not even be defined at aa (as with sin⁡xx\frac{\sin x}{x} at 00), and if it is defined, its value may differ from the limit. The figure shows a function whose value at 22 is a red dot that has nothing to do with the behaviour nearby. Second, the limit is about approach from both sides. The left-hand limit (LHL) lim⁡x→a−f(x)\lim_{x \to a^{-}}f(x) considers only x<ax < a, and the right-hand limit (RHL) lim⁡x→a+f(x)\lim_{x \to a^{+}}f(x) only x>ax > a. Then lim⁡x→af(x)\lim_{x \to a}f(x) exists   ⟺  \iff LHL and RHL both exist, are finite, and are equal. In the figure LHL =1= 1 and RHL =3= 3, so the limit does not exist. For a function defined on only one side of aa (for example x\sqrt{x} at 00), the relevant one-sided limit is taken as the limit. Evaluating one-sided limits. Substitute x=a−hx = a - h (for LHL) or x=a+hx = a + h (for RHL) with h→0+h \to 0^{+}. This is essential for piecewise functions, ∣x∣|x|, [x][x], {x}\{x\} and functions like e1/xe^{1/x}. [x][x] at an integer nn: LHL =n−1= n - 1, RHL =n= n. So lim⁡x→n[x]\lim_{x \to n}[x] does not exist, while lim⁡x→2.5[x]=2\lim_{x \to 2.5}[x] = 2. ∣x∣x\frac{|x|}{x} at 00: LHL =−1= -1, RHL =1= 1. e1/xe^{1/x} at 00: as x→0+x \to 0^{+}, 1x→+∞\frac{1}{x} \to +\infty, so e1/x→∞e^{1/x} \to \infty; as x→0−x \to 0^{-}, 1x→−∞\frac{1}{x} \to -\infty, so e1/x→0e^{1/x} \to 0. The limit does not exist. But 11+e1/x\frac{1}{1 + e^{1/x}} has LHL =1= 1 and RHL =0= 0, finite but unequal. Limits at infinity and infinite limits. lim⁡x→∞f(x)=L\lim_{x \to \infty}f(x) = L means f(x)f(x) approaches LL as xx grows without bound; the graph has a horizontal asymptote y=Ly = L. For rational functions, divide numerator and denominator by the highest power of xx in the denominator: the limit is the ratio of leading coefficients if the degrees are equal, 00 if the numerator has lower degree, and ±∞\pm\infty if it has higher degree. When we write lim⁡f(x)=∞\lim f(x) = \infty, the limit does not exist as a real number; the notation only describes how it fails. Watch the sign at −∞-\infty: lim⁡x→−∞x2+1x=−1\lim_{x \to -\infty}\frac{\sqrt{x^{2} + 1}}{x} = -1, because x2=∣x∣=−x\sqrt{x^{2}} = |x| = -x for negative xx. Existence without formulas. A limit can fail to exist without either side being infinite: sin⁡1x\sin\frac{1}{x} oscillates between −1-1 and 11 infinitely often near 00, so neither one-sided limit exists. Key Takeaways: - lim⁡x→af(x)\lim_{x \to a}f(x) ignores f(a)f(a); it depends only on values near aa. - Limit exists   ⟺  \iff LHL == RHL == a finite number. - Use x=a±hx = a \pm h, h→0+h \to 0^{+}, for one-sided limits. - [x][x] at an integer nn: LHL =n−1= n - 1, RHL =n= n. - e1/xe^{1/x}: RHL =∞= \infty, LHL =0= 0 at x=0x = 0. - At −∞-\infty, x2=−x\sqrt{x^{2}} = -x. - Rational functions at ∞\infty: compare degrees. - Pitfall: computing f(a)f(a) and calling it the limit. Brain-Teaser 1: Evaluate lim⁡x→0x[x]sin⁡∣x∣\lim_{x \to 0}\frac{x[x]}{\sin|x|} from each side, where [⋅][\cdot] is the greatest integer function. Solution: Right side: for 0<x<10 < x < 1, [x]=0[x] = 0, so the expression is 00 and RHL =0= 0. Left side: for −1<x<0-1 < x < 0, [x]=−1[x] = -1 and ∣x∣=−x|x| = -x, so the expression is −xsin⁡(−x)=−x−sin⁡x=xsin⁡x→1\frac{-x}{\sin(-x)} = \frac{-x}{-\sin x} = \frac{x}{\sin x} \to 1. So LHL =1= 1, RHL =0= 0, and the two-sided limit does not exist. Replacing [x][x] by its constant value on each side, before taking the limit, is the whole method. Brain-Teaser 2: Find lim⁡x→∞(x2+x−x)\lim_{x \to \infty}(\sqrt{x^{2} + x} - x) and lim⁡x→−∞(x2+x+x)\lim_{x \to -\infty}(\sqrt{x^{2} + x} + x). Solution: Rationalise. For x→∞x \to \infty: x2+x−x2x2+x+x=xx1+1/x+x=11+1/x+1→12\frac{x^{2} + x - x^{2}}{\sqrt{x^{2} + x} + x} = \frac{x}{x\sqrt{1 + 1/x} + x} = \frac{1}{\sqrt{1 + 1/x} + 1} \to \frac{1}{2}. For x→−∞x \to -\infty: x2+x−x2x2+x−x=x∣x∣1+1/x−x\frac{x^{2} + x - x^{2}}{\sqrt{x^{2} + x} - x} = \frac{x}{|x|\sqrt{1 + 1/x} - x}. With ∣x∣=−x|x| = -x, this is x−x(1+1/x+1)→−12\frac{x}{-x(\sqrt{1 + 1/x} + 1)} \to -\frac{1}{2}. The sign change comes entirely from x2=∣x∣\sqrt{x^{2}} = |x|, the most common error in limits at −∞-\infty. Brain-Teaser 3: Let f(x)=21/x−121/x+1f(x) = \frac{2^{1/x} - 1}{2^{1/x} + 1}. Find the one-sided limits at 00. Solution: As x→0+x \to 0^{+}, 21/x→∞2^{1/x} \to \infty. Divide numerator and denominator by 21/x2^{1/x}: 1−2−1/x1+2−1/x→1−01+0=1\frac{1 - 2^{-1/x}}{1 + 2^{-1/x}} \to \frac{1 - 0}{1 + 0} = 1. As x→0−x \to 0^{-}, 21/x→02^{1/x} \to 0, so f(x)→0−10+1=−1f(x) \to \frac{0 - 1}{0 + 1} = -1. The limit does not exist; ff behaves like a smoothed-out signum function near 00. The technique is to divide by whichever exponential is growing, so every remaining term has a finite limit.

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Limits: Meaning, One-Sided Limits and Existence — DPP

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