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

How can we check whether the limn→3+ n-1 exists? Correct Answer syms n;limit(n-1,n,3,’right’)

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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