IIT JEE , AIEEE Forum - Mathematics , Algebra

ayshwarya

Lt Sinx --(Sinx)^Sinx/ [1--Sinx +logsinx ] =?

x--> $\Pi$ /2

Lt 1-(Sinx)^Sinx/Cos²x = ?

x--> $\Pi$ /2

Suppose f(n+1) =1/2 { f(n) +9/f(n) } , n $\epsilon$ N if f(n) >0 $\forall$ n $\epsilon$ N , den Lt (f(n)) = ?

n--> $\propto$

Lt [ Cos x/m]^x =?

x-->infinity

RyuAmakusa

3) 3 .

2f(n+1) = f(n) +9/f(n) now $\mathop{\lim}\limits_{{n} \to {\infty}}2f(n+1)=\mathop{\lim}\limits_{{n} \to {\infty}}f(n)+\frac{9}{f(n)}$ if n ----> $\infty$ then n+1 -----> $\infty$ so

$\mathop{\lim}\limits_{{n} \to {\infty}}2f(n+1)=\mathop{\lim}\limits_{{n} \to {\infty}}2f(n)$ so $\mathop{\lim}\limits_{{n} \to {\infty}}2f(n)=\mathop{\lim}\limits_{{n} \to {\infty}}f(n)+\mathop{\lim}\limits_{{n} \to {\infty}}\frac{9}{f(n)}$

$\Rightarrow{(\mathop{\lim}\limits_{{n} \to {\infty}}f(n) )}^{2}=9$ $\Rightarrow\mathop{\lim}\limits_{{n} \to {\infty}}f(n)=3$

deedee

for d second q is d answer 1??

if yes i'll post d sol

ayshwarya

na deedee its 1/2 & ive got it tell me d ans 4 1st 1

ayshwarya

ya deedee ive done it using l aspital

RyuAmakusa

2) $use\;L'hospital's\;rule\;diff\;Nr\;and\;Dr$

$let\;t={sinx}{sinx}\;take\;ln\;on\;both\;sides\;lnt=sinx.ln(sinx)\;on\;diff\;we\;get\;\frac{1}{t} .\frac{dt}{dx}=cosx(1+sinx)\\\Rightarrow\frac{dt}{dx}={sinx}^{sinx}.cosx(1+lnsinx)\;now\;on\;diff\;denominator\;we\;get-2sinxcosx\;so\;the\;given\;limit\\is\mathop{\lim}\limits_{{x} \to {\frac{\pi}{2}}}=\frac{{sinx}^{sinx}(1+ln(sinx))}{2sinx}=\frac{1}{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.$

amulye

comon frnds ans d last 1 ur rite go ahead

Ask & Discuss Questions with Community & Experts