Category: 0. BLOG

Check whether lim⁡z→0zˉzlimz→0zzˉ exists in complex sense A Equals 11 B Equals −1−1 C Does not exist D Equals 00

Evaluate lim⁡z→0ez−1zlimz→0zez−1 using series idea A 00 B ee C 11 D Does not exist Explanation Using ez=1+z+z22!+⋯ez=1+z+2!z2+⋯, we get

For f(z)=z2−1z−1f(z)=z−1z2−1, what is lim⁡z→1f(z)limz→1f(z)** after simplification** A 11 B 00 C 22 D Does not exist Explanation Factor z2−1=(z−1)(z+1)z2−1=(z−1)(z+1).

While checking lim⁡z→af(z)limz→af(z), which approach best confirms path independence A Compare two different paths B Check only real axis C

Which condition must hold for lim⁡z→af(z)limz→af(z) to exist in the complex plane A Same on real axis B Same on

For f(x,y)=x2yx4+y2f(x,y)=x4+y2x2y, the limit at (0,0)(0,0) is A Does not exist B 0 C 1 D Infinity Explanation Along y=0y=0,

For f(x,y)=x2−y2x2+y2f(x,y)=x2+y2x2−y2, as (x,y)→(0,0)(x,y)→(0,0) the limit A Equals 0 B Does not exist C Equals 1 D Equals −1 Explanation

To prove lim⁡(x,y)→(0,0)f(x,y)=Llim(x,y)→(0,0)f(x,y)=L fails, it is enough to show A Function is bounded B One path exists C Two paths

When checking lim⁡(x,y)→(a,b)f(x,y)lim(x,y)→(a,b)f(x,y), the most reliable requirement is A Same value one path B Exists on x-axis C Same value

A multivariable limit exists when A One path matches B Only along axes C Values are bounded D All paths