limit

1. lim_(x->-infinity) (2^x-1/2^x)/(2^x+1/2^x)

The limit of a sum is the sum of the limits:
= lim_(x->-infinity) 2^x/(2^x+1/2^x)-lim_(x->-infinity) 1/((2^x+1/2^x) 2^x)

Simplify 1/((2^x+1/2^x) 2^x) assuming x<0 giving 1/(4^x+1):
= lim_(x->-infinity) 2^x/(2^x+1/2^x)-lim_(x->-infinity) 1/(1+4^x)

Simplify 2^x/(2^x+1/2^x) assuming x<0 giving 4^x/(4^x+1):
= lim_(x->-infinity) 4^x/(1+4^x)-lim_(x->-infinity) 1/(1+4^x)

The limit of a quotient is the quotient of the limits:
= lim_(x->-infinity) 4^x/(1+4^x)-1/(lim_(x->-infinity) (1+4^x))

The limit of a quotient is the quotient of the limits:
= (lim_(x->-infinity) 4^x)/(lim_(x->-infinity) (1+4^x))-1/(lim_(x->-infinity) (1+4^x))

The limit of a sum is the sum of the limits:
= (lim_(x->-infinity) 4^x)/(lim_(x->-infinity) 4^x+1)-1/(lim_(x->-infinity) 4^x+1)

Using the continuity of 4^x at x = -infinity write lim_(x->-infinity) 4^x as 4^(lim_(x->-infinity) x):
= 4^(lim_(x->-infinity) x)/(4^(lim_(x->-infinity) x)+1)-1/(4^(lim_(x->-infinity) x)+1)

The limit of x as x approaches -infinity is -infinity:
= -1

2. lim(x→∞) xln[1+(5/x)+(6/x^2)]
Indeterminate form of type 0·infinity. Let t = 1/x, then lim_(x->infinity) x ln(6/x^2+5/x+1) = lim_(t->0) (ln(6 t^2+5 t+1))/t:
= lim_(t->0) (ln(6 t^2+5 t+1))/t

Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(t->0) (ln(1+5 t+6 t^2))/t = lim_(t->0) (( dln(1+5 t+6 t^2))/( dt))/(( dt)/( dt)):
= lim_(t->0) (12 t+5)/(6 t^2+5 t+1)

The limit of a quotient is the quotient of the limits:
= (lim_(t->0) (12 t+5))/(lim_(t->0) (6 t^2+5 t+1))

The limit of 12 t+5 as t approaches 0 is 5:
= 5/(lim_(t->0) (6 t^2+5 t+1))

The limit of 6 t^2+5 t+1 as t approaches 0 is 1:
= 5