Subtask 1

´lim_(x -> pi/4) (sin(x) + cos(x))/(16x^2 + 1)´

Take the limit: ´lim_(x->pi/4) (cos(x)+sin(x))/(1+16 x^2)´

The limit of a quotient is the quotient of the limits: ´= (lim_(x->pi/4) (cos(x)+sin(x)))/(lim_(x->pi/4) (1+16 x^2))´

The limit of 1+16 x^2 as x approaches pi/4 is 1+pi^2: ´= (lim_(x->pi/4) (cos(x)+sin(x)))/(1+pi^2)´

The limit of a sum is the sum of the limits: ´= (lim_(x->pi/4) cos(x)+lim_(x->pi/4) sin(x))/(1+pi^2)´

The limit of cos(x) as x approaches pi/4 is 1/sqrt(2): ´= (1/sqrt(2)+lim_(x->pi/4) sin(x))/(1+pi^2)´

The limit of sin(x) as x approaches pi/4 is 1/sqrt(2): ´= sqrt(2)/(1+pi^2)´

Subtask 2

´lim_(x -> oo) (x+1)^x/x^x ´

Take the limit: ´lim_(x->infinity) (1+x)^x/x^x´

Simplify ´(1+x)^x/x^x´ assuming ´x>0´ giving ´(1+1/x)^x´: ´= lim_(x->oo) (1+1/x)^x´

Indeterminate form of type ´1^oo´. Transform using ´lim_(x->o) (1+1/x)^x = e^(lim_(x->oo) x log(1+1/x))´: ´= e^(lim_(x->oo) x log(1+1/x))´

Indeterminate form of type ´0·oo´. Let ´t = 1/x´, then ´lim_(x->oo) x log(1+1/x) = lim_(t->0) (log(1+t))/t´: ´= e^(lim_(t->0) (log(1+t))/t)´

Indeterminate form of type ´0/0´. Applying L'Hospital's rule we have, ´lim_(t->0) (log(1+t))/t = lim_(t->0) (( dlog(1+t))/( dt))/(( dt)/( dt))´: ´= e^(lim_(t->0) 1/(1+t))´

The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: ´= e^(1/(lim_(t->0) (1+t)))´ The limit of 1+t as t approaches 0 is 1:

´= e´

Subtask 3

´lim_(x -> 1) (x^3 - 6x^2 + 11x - 6)/(x^2 - 5x + 4)´

Take the limit: ´lim_(x->1) (-6+11 x-6 x^2+x^3)/(4-5 x+x^2)´

Factor the numerator and denominator: ´= lim_(x->1) ((-1+x) (6-5 x+x^2))/((-1+x) (-4+x))´

Cancel terms, assuming ´-1+x!=0´: ´= lim_(x->1) (6-5 x+x^2)/(-4+x)´

The limit of a quotient is the quotient of the limits: ´= (lim_(x->1) (6-5 x+x^2))/(lim_(x->1) (-4+x))´

The limit of ´-4+x´ as ´x´ approaches ´1´ is ´-3´: ´= -1/3 (lim_(x->1) (6-5 x+x^2))´

The limit of ´6-5 x+x^2´ as ´x´ approaches ´1´ is ´2´: ´= -2/3´