Question

How can we check whether the limn→3+ n-1 exists?

(a) limit(n-1,n,3)

(b) syms n;limit(n-1,n,3,’right’)

(c) syms n;limit(n-1,3,’right’)

(d) syms n;limit(n-1,3,n,’right’)

This question was posed to me during an interview for a job.

I want to ask this question from Limits topic in division Beyond the Basics of MATLAB

Answer 1

Correct choice is (b) syms n;limit(n-1,n,3,’right’)

Easiest explanation: We have to give our symbolic argument before our limit command. Hence, syms n;limit(n-1,n,3,’right’) has the right order, and not syms n;limit(n-1,3,n,’right’). syms n;limit(n-1,n,3,’right’) is not same as limit(n-1,n,3) since there we are calculating the value of the function at the limiting value, 3, but we need to find it at the right-hand side of the limiting value- plus, limit(n-1,n,3) does not declare n as symbolic- this will give an error. syms n;limit(n-1,3,’right’) will give an error since the symbolic argument is not provided into the command.