Continuity

4 Marks Questions

1. Find the value of k, so that the function f defined below, is continuous at <math>x = 0</math>, where

<math display="block">f(x) = \begin{cases} \left(\frac{1 - \cos 4x}{8x^2}\right), & \text{if } x \neq 0\\ k, & \text{if } x = 0 \end{cases}</math> All India 2014C

Given function is

<math display="block">f(x) = \begin{cases} \left(\frac{1 - \cos 4x}{8x^2}\right), & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}</math>

Also, given f(x) is continuous at <math>x = 0</math>.

<math>(LHL)_{x=0} = (RHL)_{x=0} = f(0)</math> ...(i)

Now, LHL = <math>\lim_{x \to \infty} f(x)</math>

<math>x \rightarrow 0^{-}</math>

<math>=\lim_{x\to 0^{-}}\frac{1-\cos 4x}{8x^2}</math> <math display="block">= \lim_{h \to 0} \frac{1 - \cos(-4h)}{8h^2}</math>

[put <math>x = 0 - h = -h</math>, when <math>x \to 0</math>, <math>h \to 0</math>](1) <math display="block">= \lim_{h \to 0} \frac{1 - \cos 4h}{8h^2}</math>

<math display="block">= \lim_{h \to 0} \frac{2 \sin^2 2h}{8h^2}</math>

<math display="block">\lim_{h \to 0} \frac{\sin^2 2h}{4h^2} = \lim_{h \to 0} \left(\frac{\sin 2h}{2h}\right)^2 = 1 </math> (1)

At <math>x = 0</math>, <math>f(0) = k</math>

Now, from Eq. (i), we have

<math>LHL = f(0)</math>

<math>1 \cdot 1 = k \Rightarrow k = 1</math> (1) <math>\Rightarrow</math>

Hence, for <math>k = 1</math>, the given function f(x) is continuous at <math>x = 0</math>. (1)

1. If <math>f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & \text{when } x < 0 \\ \frac{x}{\sqrt{x}}, & \text{when } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}}, & \text{when } x > 0 \end{cases}</math>

and f is continuous at <math>x = 0</math>, then find the value of a. Delhi 2013C

Given,

<math>f(x) =</math> <math display="block">\frac{1-\cos 4x}{x^2}, \text{ when } x<0</math>

<math display="block">a, \text{ when } x=0</math>

<math display="block">\frac{\sqrt{x}}{(\sqrt{16+\sqrt{x}})-4}, \text{ when } x>0</math>

Since, f(x) is continuous at <math>x = 0</math>.

<math display="block">\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = f(0)</math> <math>\Rightarrow \lim_{x \to 0^{-}}</math> <math display="block">\frac{1-\cos 4x}{x^2}</math> <math display="block">= \lim_{x \to 0^+} \frac{\sqrt{x}}{(\sqrt{16 + \sqrt{x}}) - 4} = f(0) \quad (1)</math>