Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question