Question

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

Answer 1

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

Answer 2

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