Author: Study Clue

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

For f(x)=x4−4x2f(x)=x4−4×2, which xx-values are inflexion candidates from f′′(x)=0f′′(x)=0? A x=±13x=±31 B x=±1x=±1 C x=±2x=±2 D x=±23x=±32 Explanation f′′(x)=12×2−8=4(3×2−2)f′′(x)=12×2−8=4(3×2−2). Setting

For f(x)=x3−3xf(x)=x3−3x, the point of inflexion occurs at A x=0x=0 B x=1x=1 C x=−1x=−1 D x=3x=3 Explanation f′′(x)=6xf′′(x)=6x. Setting f′′(x)=0f′′(x)=0

For a twice-differentiable function, the condition “f′(x)f′(x) increasing on an interval” directly implies A f′′(x)≤0f′′(x)≤0 B f′(x)=0f′(x)=0 C f′′(x)≥0f′′(x)≥0 D

While sketching y=f(x)y=f(x), the interval where f′′(x)>0f′′(x)>0 is mainly used to show A Downward bending B No curve change C

When a function is concave up on an interval, what is true about f′′(x)f′′(x) there? A f′′(x)0 D f′(x)=0f′(x)=0 Explanation