Question #761

evaluate (a) lim_(x→0)(√(x^2+x^4 )*sin(π/x));
(b) lim_(x→0+)(√(x) (1+sin^2(π/x)));

Expert's answer

A. At x->0 sin(π/x) has no limit, but it keeps oscillating from -1 to +1 and back, so it's values are limited [-1,1]

lim_(x->0) √(x^2+x^4) = 0

the product of 0 and limited function would be equal to zero.

B. The function sin^2(pi/x) is limited in the domain [0,1], so (1+ sin^2(pi/x)) has the values from 1 to 2.

lim_(x->0) √(x) = 0;

the product of 0 and the limited function would be equal to zero

lim_(x->0) √(x^2+x^4) = 0

the product of 0 and limited function would be equal to zero.

B. The function sin^2(pi/x) is limited in the domain [0,1], so (1+ sin^2(pi/x)) has the values from 1 to 2.

lim_(x->0) √(x) = 0;

the product of 0 and the limited function would be equal to zero

## Comments

## Leave a comment