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  • <img alt="= \lim_{x\rightarrow 0} |x(x-1)|^{[\cos 2 x ]}\\\\" src="https://learn.careers360.com/latex-image/?%3D%20%5Clim_%7Bx%5Crightarrow%200%7D%20%7Cx%28x-1%29%7C%5E%7B%5B%5Ccos%202%20x%20%5D%7D

= \lim_{x\rightarrow 0} |x(x-1)|^{[\cos 2 x ]}\\\\where [.] denotes greatest integer function, is equal to

Option: 1

1 


Option: 2

0


Option: 3

e


Option: 4

does not exist 


Answers (1)

 

Right hand limit -

The right hand limit of  f(x) as 'x' tends to 'a' exists and is equal to  l1, if as  'x'  approaches 'a' through values greater than 'a'.
 

so\:\lim_{x\rightarrow a^{+}}f(x)=l_{1} 

- wherein

where  a+ means  a+h  &  h → 0.  Therefore f(a+h).

 

 

 

Left hand Limit -

The left hand limit of f(x) as  'x'  tends to 'a' exists and  is equal to l2, if as 'x'  approaches 'a'  through values less than 'a'.


\lim_{x\rightarrow a^{-}}f(x)= l_{2}

- wherein

Where  a-   means ( a - h ) & h\rightarrow 0. Therefore, f(a-h).

 

 

 

 

RHL =\lim_{h \rightarrow 0 ^+ } | h ( h-1 )| ^{[\cos 2h ]}= \lim_{h \rightarrow 0 ^+ } h ( h-1 )) ^0 = 1 \\\\ LHL = \lim_{h \rightarrow 0 ^- } | -h ( -h-1 )| ^{[\cos 2h ]} = \lim_{h \rightarrow 0 ^- } ( h ( 1+h )) ^0 = 1 

required limit = 1 

Posted by

Ramraj Saini

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