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  • Help me answer: - Limit , continuity and differentiability - JEE Main-4

\lim_{x\rightarrow 0}\left ( \left [ \frac{m \sin x}{x}\right ]+ \left [ \frac{n \tan x }{x} \right ] \right ) equals 

(m> 0,n< 0,m,m\epsilon I,[.]= G.I.F)

  • Option 1)

    m+n-2

  • Option 2)

    m+n-1

  • Option 3)

    m+n

  • Option 4)

    m+n+1

 

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

 

 \frac{m \sin x}{x}\rightarrow m  but less than m so \left [ \frac{m \sin x}{x} \right ]=(m-1 )  

\frac{n\tan x}{x}\rightarrow n but less than n , so \left [ \frac{n \tan x}{x} \right ]=(n-1 )

\therefore (m-1)+(n-1) = (m+n-2)

 

 

 

 


Option 1)

m+n-2

Option 2)

m+n-1

Option 3)

m+n

Option 4)

m+n+1

Posted by

Himanshu

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