Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

For what value of k, is the following function continuous at x = 0? $ f(x) = \left\{ \begin{array}{l l}\large\frac{1-\cos4x}{8x^2}, & \quad if { x \neq 0 } \\ k, & \quad if { x = 0 } \end{array} \right. $

Can you answer this question?

1 Answer

0 votes
  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
$ f(x) = \left\{ \begin{array}{l l}\large\frac{1-\cos4x}{8x^2}, & \quad if { x \neq 0 } \\ k, & \quad if { x = 0 } \end{array} \right. $
Consider the LHL
$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}\large\frac{1-\cos 4x}{8x^2}$
$1-\cos 4x=2\sin^22x$
$\therefore \lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}\large\frac{2\sin^22x}{8x^2}$
$\qquad\qquad\;\;=\lim\limits_{x\to 0^-}\large\frac{\sin^22x}{4x^2}$
$\qquad\qquad\;\;=\lim\limits_{x\to 0^-}\big(\large\frac{\sin 2x}{2x}\big)^2$
But we know $\lim\limits_{x\to 0}\large\frac{\sin \theta}{\theta}$$=1$
Hence $\lim\limits_{x\to 0^-}\big(\large\frac{\sin 2x}{2x}\big)^2$$=1$
Step 2:
Consider the RHL
$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^+}k$
It is given that the function is continuous at $x=0$
$\therefore$ LHL =RHL
$\Rightarrow 1=k$
Hence the value of $k=1$
answered Oct 3, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App