# 2013 solution - University of Toronto Scarborough...

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University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT B41H 2013/2014 Term Test Solutions 1. (a) From the lecture notes we have Let f : U R n R k be a given function. We say that f is differentiable at a U if the partial derivatives of f exist at a and if lim x a bardbl f ( x ) f ( a ) Df ( a ) ( x a ) bardbl bardbl x a bardbl = 0 , where Df ( a ) is the k × n matrix parenleftbigg ∂f i ∂x j parenrightbigg evaluated at a . Df ( a ) is called the derivative of f at a . (b) From the lecture notes we have Chain Rule . Let f : U R n R m and g : V R m R k be given functions such that f [ U ] V so that g f is defined. Let a R n and b = f ( a ) R m . If f is differentiable at a and g is differentiable at b , then g f is differentiable at a and D ( g f ) ( a ) = [ Dg ( b )] [ Df ( a )] . 2. (a) (i) lim ( x,y ) (0 , 0) xy 2 x 2 + y 2 . For ( x,y ) negationslash = (0 , 0), we have 0 vextendsingle vextendsingle vextendsingle vextendsingle y 2 x 2 + y 2 vextendsingle vextendsingle vextendsingle vextendsingle 1 because x 2 0. Now multiplying by | x | gives 0 vextendsingle vextendsingle vextendsingle vextendsingle xy 2 x 2 + y 2 vextendsingle vextendsingle vextendsingle vextendsingle ≤ | x | . Hence 0 lim ( x,y ) (0 , 0) vextendsingle vextendsingle vextendsingle vextendsingle xy 2 x 2 + y 2 vextendsingle vextendsingle vextendsingle vextendsingle lim ( x,y ) (0 , 0) | x | . Since lim ( x,y ) (0 , 0) | x | = 0 , lim ( x,y ) (0 , 0) vextendsingle vextendsingle vextendsingle vextendsingle xy 2 x 2 + y 2 vextendsingle vextendsingle vextendsingle vextendsingle = 0 by the Squeeze Theorem. Hence lim ( x,y ) (0 , 0) xy 2 x 2 + y 2 = 0 as well. (Other approaches are possible.) (ii) lim ( x,y ) (0 , 0) x 3 y xy 3 x 4 + y 4 . Evaluating along the line y = 0, the limit reduces to lim x 0 0 x 4 = 0. On the other hand, evaluating along the line y = 2 x , the limit reduces to lim x 0 2 x 4 8 x 4 2 x 4 = lim x 0 3 x 4 x 4 = 3. Since 3 negationslash = 0, the limit does not exist.
MATB41H Term Test Solutions page 2 (b) For f to be continuous at (0 , 0) we need lim ( x,y ) (0 , 0) f ( x,y ) = 0 = f (0 , 0).