Question

Calculate the next limit, if it doesn’t exist, then prove it.

2 y (b) lim (x,y)→(0,0) sin' y + ln(1 + x2)

Answer 1

lim_(x, y) rightarrow (0, 0) (x^2 + y^2/sin^2y + ln(1 + x^2)) lim_x rightarrow 0 y rightarrow 0 (x^2 + y^2/sin^2 y + ln (1 + x^2)) = lim_y rightarrow 0 (0 + y^2/sin^2 y + ln (1 + 0^2)) = lim_y rightarrow 0 (y^2/sin^2 y) = lim_y rightarrow 0 (2y/2 siny cosy) = lim_y rightarrow 0 (2y/sin 2y) = l im_y rightarrow 0 (2/2 cos 2y) = lim_y rightarrow 0 (1/cos 2y) = 1 lim_y rightarrow 0 x rightarrow 0 (x^2 + y^2/sin^2 y + ln (1 + x^2)) = lim_x rightarrow 0 (0 + x^2/0 + ln(1 + x^2)) = lim_x rightarrow 0 (2x/2/1 + x^2) = lim_x rightarrow 0 (1 + x^2) = 0 put y = mx lim_(x, y) rightarrow (0, 0) (x^2 + y^2/sin^2y + ln(1 + x^2)) = lim_x rightarrow 0 (x^2 + m^2x^2/sin^2(mx) + ln(1 + x^2)) = lim_x rightarrow 0 (2x(1 + m^2)/2m sin(mx) cos(mx) + 2x/(1 + x^2)) = lim_x rightarrow 0 (2x(1 + m^2)/m sin(2mx) + 2x/(1 + x^2)) = lim_x rightarrow 0 (2(1 + m^2)/2m^2 cos(2mx) + (1 + x^2)2 - 2x(2x)/(1 + x^2)^2)) = 1