Lots of Limits

Calculus Level 4

Which of the Following Limits Does exist.

A: lim ( x , y ) ( 0 , 0 ) x y x 2 + y 2 \quad \large\displaystyle \lim_{(x,y)\to (0,0)} \dfrac{xy}{x^2 + y^2}

B: lim ( x , y ) ( 0 , 0 ) x + y x 2 + y 2 \quad \large\displaystyle \lim_{(x,y)\to (0,0)} \dfrac{x + \sqrt y}{x^2 + y^2}

C: lim ( x , y ) ( 0 , 0 ) x 3 y x 6 + y 2 \quad \large\displaystyle \lim_{(x,y)\to (0,0)} \dfrac{x^3y}{x^6 + y^2}

D: lim ( x , y ) ( 0 , 1 ) ( y 1 ) tan 2 x x 2 ( y 2 1 ) \quad \large\displaystyle \lim_{(x,y)\to (0,1)} \dfrac{(y - 1)\tan^2x}{x^2(y^2 - 1)}

D B A C

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1 solution

For the 1st one .... approaching ( 0 , 0 ) (0,0) through y = m x y=mx .

we get the limit equals m 1 + m 2 \displaystyle \frac{m}{1+m^2} which is different for different values of m .

Hence limit Does not exist.

for 2:-

approaching ( 0 , 0 ) (0,0) through y = x 2 \displaystyle y=x^2 , we see that the right hand limit tends to positive \infty and left hand limit tends to negative \infty . Hence limit does not exist.

for 3 :-

putting y = m x 3 \displaystyle y = mx^3 we see that the limit equals m 1 + m 3 \displaystyle \frac{m}{1+m^3} which is different for different values of m . Hence limit does not exist.

for 4:-

we see that the limit equals ( lim y 1 1 y + 1 ) ( lim x 0 ( tan ( x ) x ) 2 ) \displaystyle \left(\lim_{y\to 1}\frac{1}{y+1}\right)\left(\lim_{x\to 0}\left(\frac{\tan(x)}{x}\right)^{2}\right) .

which equals 1 2 \displaystyle \frac{1}{2} .

Hence D D is the only correct option

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