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  • Can someone explain - Let denote the greatest integer less than or equal to. Then : : - Limit , continuity and differentiability - JEE Main

 

Let \begin{bmatrix} x \end{bmatrix}  denote the greatest integer less than or  equal to x . Then :

\lim_{x\rightarrow 0}\frac{\tan (\pi \sin ^{2}x)+(\left | x \right |-\sin (x\begin{bmatrix} x \end{bmatrix}))}{x^{2}}^{2} :

  • Option 1)

     

    equals \pi+1

  • Option 2)

     

    equals 0

  • Option 3)

     

    does not exist

  • Option 4)

     

    equals \pi

Answers (1)

best_answer

 

Evalution of Trigonometric limit -

\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1

\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1

put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0

Then\:it\:comes

\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1

\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and

\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=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).

 

 

\lim_{x\rightarrow 0}\frac{\tan \left ( \pi \sin ^{2}x \right )+\left ( \left | x \right | -\sin \left ( x\left [ x \right ] \right )\right )^{2}}{x^{2}}

=\lim_{x\rightarrow 0}\left [ \frac{\tan \left ( x\sin ^{2}x \right )}{x\sin ^{2}x} \times \simeq \left ( \frac{\sin x}{x} \right )^{2}+\frac{x^{2}}{x^{2}}-\frac{2\left | x \right |\sin \left ( x\left [ x \right ] \right )}{x^{2}}+\frac{\sin ^{2}\left ( x\left [ x \right ] \right )}{x^{2}}\right ]

RHS, \lim\: \: \: x\rightarrow 0+,      \left | x \right |=x  and   \left [ x \right ]=0

So, RHS = \left ( \pi +1 \right )

LHS, \lim\: \: \: x\rightarrow 0-,    \left | x \right |=-x     and    \left [ x \right ]=-1

So, LHS =\left ( \pi +1-2+1 \right )=\pi

Hence limit does not exist 

 

 


Option 1)

 

equals \pi+1

Option 2)

 

equals 0

Option 3)

 

does not exist

Option 4)

 

equals \pi

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