Question

Given that limit exists find \(lt_{(x,y,z)\rightarrow(-2,-2,-2)}\frac{sin((x+2)(y+5)(z+1))}{(x+2)(y+7)}\)

(a) 1

(b) ^3⁄5

(c) ^1⁄2

(d) 0

I had been asked this question during an online interview.

The origin of the question is Limits and Derivatives of Several Variables in chapter Partial Differentiation of Engineering Mathematics

Answer 1

Right answer is (b) ^3⁄5

Explanation: Given that limit exists we can parameterize the curve

Put x = t : y = t : z = t

\(lt_{t\rightarrow -2}\frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+7)}\)

\(lt_{t\rightarrow -2}\frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+5)(t+1)}\times lt_{t\rightarrow -2}\frac{(t+5)(t+1)}{(t+7)}\)

\((1)\times \frac{(-2+5)(-2+1)}{(-2+7)}\)

=\(\frac{(3).(1)}{(5)}=\frac{3}{5}\)