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mastermind890

_{[x ]}

_{[ 0]}(log cot x)^{tan x}

rini

anser is zero.

log(cot x)^{tan x}

= log cot x/cot x

apply l'hospitals rool...

1/cot x (-cosec² x)/cosec²x

=1/cotx=tan x

=0

hence ans should be 0.

plz raTE....

umang

Hey rini !

You cant apply LH rule , as it is applied only in 0/0 and

/

form .

So , u r wrong in this step .

and , u hav misread the question . Its (log cot x)^{tan x} and not

log(cot x)^{tan x .}

^{I hope u understood !!!}

krzme

1) f(lt g(x)) = lt f(g(x));

hence apply log to the given ques & solve to get ans say y.

then y=log(actual ans)

hence e^y=actual ans.

2) log(infinity)=infiniy (rem log is an incresing func?!)

3) follow rini to get 0.

the actual ans hence is: e^0=1.

phanindraramesh

_{[x ]}

_{[ 0]} [log(cotx)]^tanx

_{=[x ]}

_{[ 0]} [1+log(cotx)-1]^tanx(let me replace_{[x ]}

_{[ 0^]}with q for some time)

=e^{q [log(cotx) - 1]tanx}

{as x->a if f(x)=o and g(x)=0 orf(x)= infinity and g(x)=infinity

then

_{[x ][a ]^{[1+f(x)]^g(X) is e to the power of f(x)g(x) with limit}}}

=e^{q log(cotx)-1_cotx}

=e^{q - cosec^2x/cotx_-cosec^2x}[by using l'hoosptal rule in the power]

=e^{q 1/cotx}

=e^1/infinity

=e⁰=1

I think hte answer is clear

Bye..........................

akku

it is a

^0 form.

y=_{[x ]}

_{[0 ]} log cotx ^tanx

log y=_{x ][0 ]} (log logcotx )*tanx=_{x ][0 ]} (log logcotx )/cotx (0/0) form

Applying L'Hospital

log y=

_{[x ][0 ]} 1/logcotx*1/cotx*-cosec^2x/-cosec^2x=_{x ][0 ]} tanx/logcotx= 0/

=0

logy=0 => y=e^0=1 ans

umang

All of u hav given the perfect solution !!!

Thanks !!!

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