Question

Given that limit exists find \(lt_{(x,y,z)\rightarrow(-1,-1,-1)}\frac{tan((x-1)(y-2)(z-3))}{(x-1)(y-6)(z+7)}\)

(a) 1

(b) ^1⁄2

(c) ^1⁄7

(d) ^2⁄7

This question was addressed to me in a job interview.

Query is from Limits and Derivatives of Several Variables in division Partial Differentiation of Engineering Mathematics

Answer 1

Correct choice is (d) ^2⁄7

Explanation: We can parameterize the curve by

x=y=z=t

\(lt_{t\rightarrow -1}\frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-6)(t+7)}\)

\(lt_{t\rightarrow -1}\frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-2)(t-3)}\times lt_{t\rightarrow -1}\frac{(t-2)(t-3)}{(t-6)(t+7)} \)

=\(\frac{(-1-2)(-1-3)}{(-1-6)(-1+7)}=\frac{(3)(4)}{(7)(6)}\)

=\(\frac{12}{42}=\frac{2}{7}\)