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  • Please help! - Limit , continuity and differentiability - JEE Main-8

\lim_{x\rightarrow 0} \left [ \frac{m \sin x}{x} \right ] equals 

  • Option 1)

    m when m is positive

  • Option 2)

    m-1 when m is positive

  • Option 3)

    m-1 when m is negative

  • Option 4)

    m+1 when m is negative

 

Answers (1)

best_answer

As we have learned

Condition for Trigonometric limit -

\lim_{x\rightarrow 0}\:\:\:[\frac{sinx}{x}]=0

\lim_{x\rightarrow 0}\:\:\:[\frac{tanx}{x}]=1

[\lim_{x\rightarrow 0}\:\frac{sinx}{x}]=[\lim_{x\rightarrow 0}\:\frac{tanx}{x}]=1

- wherein

Where  [ . ] is greater integer function

 

 when m is positive , then \frac{m\sin x}{x}  approaches 'm' but will be less than 'm' 

[\frac{m\sin x}{x}]=(m-1 )

when  m is negative , then  \frac{m\sin x}{x} approachers m but will be greater than m so 

[\frac{m\sin x}{x}]=m

 

 

 

 

 


Option 1)

m when m is positive

Option 2)

m-1 when m is positive

Option 3)

m-1 when m is negative

Option 4)

m+1 when m is negative

Posted by

Himanshu

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