Find `lim_(n->0) (e^n-e^-n)`

If A=[{:(,1,a),(,0,1):}] then find lim_(n-oo) (1)/(n)A^(n)

If f(x) = lim_(n->oo) tan^(-1) (4n^2(1-cos(x/n))) and g(x) = lim_(n->oo) n^2/2 ln cos(2x/n) then lim_(x->0) (e^(-2g(x)) -e^(f(x)))/(x^6) equals

Evaluate: lim_(n->0)(e^(sinx)-(1+sinx))/({tan^(-1)(sinx)}^2)

Find lim_( x to 0) (sin x^(n))/((sin x)^(m)) " where" , m , n in Z^(+) equal

evaluate lim_(n->oo)((e^n)/pi)^(1/ n)

evaluate lim_(n->oo)((e^n)/pi)^(1/ n)

Evaluate the following limit: (lim)_(x->0)(e^(t a n x)-1)/(t a n x)

If f(n+1)=1/2{f(n)+9/(f(n))},n in N , and f(n)>0 for all n in N , then find lim_(n->oo)f(n)

f'(0) = lim_(n->oo) nf(1/n) and f(0)=0 Using this, find lim_(n->oo)((n+1)(2/pi)cos^(- 1)(1/n)-n)),|cos^(-1)1/n|

Find the value of n, if lim_(xto2) (x^(n)-2^(n))/(x-2)=80 , n in N .