Math, asked by molinadube, 1 year ago

lim sin x ^ tan x , as x tends to pi/2

Answers

Answered by abhi178
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

Anonymous: Wow!
Answered by Anonymous
8
Heya user,

 \lim_{x \to \ \pi /2}} sinx^{tanx }

Now, Consider sin x = a, tanx = x, then, from the expansion rule for a^x, we have : 

----> \lim_{x \to \ \pi /2}} sinx^{tanx }
 \lim_{x \to \pi/2} [ 1 + tanx(log sinx)+(tan^2x)/2!(log sinx)^2+... - (i)

Now, as {x \to \pi/2} --\ \textgreater \  {tan x \to \infty} and by rule,

-->  \lim_{n \to \infty} n * 0 = 0

So, (i) changes to ---> 

=  \lim_{tanx \to \infty} [ 1 + tanx * 0 + ...] = 1 <-- The desired result

Anonymous: :) <--- I avoided the L-Hospital Rule
abhi178: very nicely explained. keep it
Anonymous: HEHE!! I'm new to limits
Anonymous: Skipped school for the two day study to be able to answer that Qn
abhi178: thats great .
Nikki57: Perfect :)
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