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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»

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.


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