Simple Limit! 1
Find the value of
x → 0 lim x 2 sin ( π cos 2 x )
The answer is 3.14.
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2 solutions
How is x 2 s i n ( π s i n 2 x ) = x 2 π s i n 2 x
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k → 0 lim k s i n ( k ) = 1
Short trick
When the angle of sin tends to zero you can replace it by angle
k → 0 lim s i n ( k ) = k → 0 lim k
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The statement of lim k → 0 sin k = k does not make mathematical sense, because k is a variable, and hence is meaningless on the RHS.
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@Calvin Lin – I mean to say limit is on both sides, let me edit to clear what I mean to say
What I mean by this is when it tense to 'zero' sin (k) = k
JEE MAIN 2014 QUESTION CAN BE DONE BY L'HOPITAL'S RULE TOO
For all those who are saying why
x 2 sin ( π sin 2 x ) = x 2 π sin 2 x
We know that lim z → 0 z sin z = 1
Multiply and divide by π s i n 2 x
Now
lim x → 0 π sin 2 x sin ( π sin 2 x ) = 1
We will finally get
lim x → 0 x 2 π s i n 2 x
Again
lim x → 0 x sin x = 1
Finally we will finally get answer π
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You cannot just swap the variable z with the constant x . I think what you are trying to say is "Substitute z with π sin 2 x , which tends to 0."
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Yes :) i used 'z' because we already have 'x' in the problem so as no one get confused
Let 0=0.000001 so the result won't be indeterminate..
x → 0 lim x 2 sin ( π ( 1 − sin 2 x ) )
x → 0 lim x 2 s i n ( π s i n 2 x )
x → 0 lim x 2 π s i n 2 x
= π