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  • Confused! kindly explain, - Limit , continuity and differentiability - JEE Main-8

\lim_{x\rightarrow 0}(\left [ \frac{-3 \sin x}{x} \right ] +\left [ \frac{-3 \tan x }{x} \right ])  equals 

  • Option 1)

    -4

  • Option 2)

    -5

  • Option 3)

    -6

  • Option 4)

    -7

 

Answers (1)

best_answer

As we have learned

Condition of Trigonometric Limits -

\lim_{x\rightarrow 0}\:\:\:\frac{sinx}{x}=1\:(it\:is\:less\:than\:1)


\lim_{x\rightarrow 0}\:\:\:\frac{tanx}{x}=1\:(it\:is\:more\:than\:1)

- wherein

because\:\:\:\:\:\frac{sinx}{x}<1

                    
                      \frac{tanx}{x}>1

 

 \frac{-3\sin x}{x} will approach -3 but will be greater than -3\Rightarrow \left [ \frac{-3\sin x}{x} \right ]= -3 

and \left [ \frac{-3\tan x}{x} \right ] will also approach -3 but will be less than \Rightarrow \left [ \frac{-3\tan x}{x} \right ] = -4

-3+(-4)=-7

 

 

 


Option 1)

-4

Option 2)

-5

Option 3)

-6

Option 4)

-7

Posted by

Himanshu

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