lim sin x ^ tan x , as x tends to pi/2
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Answered by
39
Lim(x→π/2) sinx^tanx
first of all we have to know that which form of limit in this question. put x = π/2
0^∞ is the form of limit .
now,
Lim(x→π/2) e^{tanx.logsinx}
Lim(x→π/2)e^{logsinx}/cotx
e^[Lim(x→π/2){logsinx}/{cotx}]
now check the form of limit . put x = π/2
so,logsinπ/2/cotπ/2 = 0/0
hence, now we can use L - Hospital rule .
so, e^[Lim(x→π/2){cosx/sinx}/{-cosec²x}]
= e^[Lim(x→π/2){cosx/-sinx)}
now, put x = π/2
we get ,
= e^{0/1} = e^0 = 1
first of all we have to know that which form of limit in this question. put x = π/2
0^∞ is the form of limit .
now,
Lim(x→π/2) e^{tanx.logsinx}
Lim(x→π/2)e^{logsinx}/cotx
e^[Lim(x→π/2){logsinx}/{cotx}]
now check the form of limit . put x = π/2
so,logsinπ/2/cotπ/2 = 0/0
hence, now we can use L - Hospital rule .
so, e^[Lim(x→π/2){cosx/sinx}/{-cosec²x}]
= e^[Lim(x→π/2){cosx/-sinx)}
now, put x = π/2
we get ,
= e^{0/1} = e^0 = 1
Anonymous:
Wow!
Answered by
8
Heya user,
Now, Consider sin x = a, tanx = x, then, from the expansion rule for , we have :
---->
= - (i)
Now, as and by rule,
-->
So, (i) changes to --->
= 1 <-- The desired result
Now, Consider sin x = a, tanx = x, then, from the expansion rule for , we have :
---->
= - (i)
Now, as and by rule,
-->
So, (i) changes to --->
= 1 <-- The desired result
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