A) exists and is equals \[[x]\]
B) exists and is equals \[\underset{x\to 0}{\mathop{\lim }}\,f[x]\]
C) does not exist because \[f(x)=|px-q|+r|x|\]
D) does not exist because left hand limit is not equal to right hand limit
Correct Answer: D
Solution :
\[f(1+)=\underset{h\to 0}{\mathop{\lim }}\,f(1+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+\cos 2h}}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{2}\sin \,h}{h}=\sqrt{2}\] \[f(1-)=\underset{h\to 0}{\mathop{\lim }}\,f(1-h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos (-2h)}}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\sqrt{2}\frac{\sin \,h}{-h}=-\sqrt{2}\] \[\because \] \[LHS\ne RHL\] \[\therefore \] limit does not exist
You need to login to perform this action.
You will be redirected in
3 sec
