"Subtask 1\n\n´lim_(x -> pi/4) (sin(x) + cos(x))/(16x^2 + 1)´\n\nTake the limit:\n´lim_(x->pi/4) (cos(x)+sin(x))/(1+16 x^2)´\n\nThe limit of a quotient is the quotient of the limits:\n´= (lim\(x->pi/4) (cos(x)+sin(x)))/(lim\(x->pi/4) (1+16 x^2))´\n \nThe limit of 1+16 x^2 as x approaches pi/4 is 1+pi^2:\n´= (lim\(x->pi/4) (cos(x)+sin(x)))/(1+pi^2)´\n \nThe limit of a sum is the sum of the limits:\n´= (lim\(x->pi/4) cos(x)+lim\(x->pi/4) sin(x))/(1+pi^2)´\n \nThe limit of cos(x) as x approaches pi/4 is 1/sqrt(2):\n´= (1/sqrt(2)+lim\(x->pi/4) sin(x))/(1+pi^2)´\n \nThe limit of sin(x) as x approaches pi/4 is 1/sqrt(2):\n´= sqrt(2)/(1+pi^2)´\n\n\nSubtask 2\n\n´lim\(x -> oo) (x+1)^x/x^x ´\n\nTake the limit:\n´lim\(x->infinity) (1+x)^x/x^x´\n\nSimplify ´(1+x)^x/x^x´ assuming ´x>0´ giving ´(1+1/x)^x´:\n´= lim\(x->oo) (1+1/x)^x´\n\nIndeterminate form of type ´1^oo´. Transform using ´lim\(x->o) (1+1/x)^x = e^(lim\(x->oo) x log(1+1/x))´:\n´= e^(lim\(x->oo) x log(1+1/x))´\n\nIndeterminate 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´:\n´= e^(lim\(t->0) (log(1+t))/t)´\n\nIndeterminate 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))´:\n´= e^(lim\(t->0) 1/(1+t))´\n\nThe limit of a quotient is the quotient of the limits:\nThe limit of a constant is the constant:\n´= e^(1/(lim\(t->0) (1+t)))´\nThe limit of 1+t as t approaches 0 is 1:\n\n´= e´\n\n\nSubtask 3\n\n´lim\(x -> 1) (x^3 - 6x^2 + 11x - 6)/(x^2 - 5x + 4)´\n\nTake the limit:\n´lim\(x->1) (-6+11 x-6 x^2+x^3)/(4-5 x+x^2)´\n\nFactor the numerator and denominator:\n´= lim\(x->1) ((-1+x) (6-5 x+x^2))/((-1+x) (-4+x))´\n\nCancel terms, assuming ´-1+x!=0´:\n´= lim\(x->1) (6-5 x+x^2)/(-4+x)´\n\nThe limit of a quotient is the quotient of the limits:\n´= (lim\(x->1) (6-5 x+x^2))/(lim_(x->1) (-4+x))´\n\nThe limit of ´-4+x´ as ´x´ approaches ´1´ is ´-3´:\n´= -1/3 (lim\_(x->1) (6-5 x+x^2))´\n\nThe limit of ´6-5 x+x^2´ as ´x´ approaches ´1´ is ´2´:\n´= -2/3´"