What Is the Limit of n*sin(1/n^2) as n Approaches Infinity?

  • Thread starter dickshant
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In summary, the conversation is about finding the limit as n approaches infinity of two different expressions involving square roots and trigonometric functions. The person asking for help is having trouble understanding the question and the symbols being used, but the other person suggests a possible solution for each expression.
  • #1
dickshant
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Hello, can anyone help me with these questions... At least one of them please... thank you

1 Find the limit as → ∞, if it exists, of:(i) √︀( ( + 1)) − 2 , (ii) sin(1/(^2))
I really cannot understand how to do this.:confused:..thanks for your help:):)
 
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  • #2


You seem to be using a number of "special symbols". All my viewer shows is a square for each.
 
  • #3


ok...i will reformulate the question
find the limit as n tends to infinity, if it exists of
1. root(n(n+1))-2n

2. n*sin(1/n^2)
Thanks...:)
 
  • #4


For the first one, multiply out what you have in the square root and pull out n2 from the square root.
For the second, you might try the substitution u = 1/n
 

What is the purpose of solving the limit of n*sin(1/n^2)?

The purpose of solving this limit is to understand the behavior of the function as n approaches infinity. It can also help in finding the maximum and minimum values of the function and determining its continuity.

What is the general approach to solving this type of limit?

The general approach is to simplify the expression by factoring out the dominant term in the denominator. Then, use algebraic manipulation and the properties of limits to evaluate the limit.

What is the value of this limit?

The value of this limit is 0.

Why is the limit of n*sin(1/n^2) equal to 0?

This limit is equal to 0 because as n approaches infinity, 1/n^2 approaches 0, and the sine function of 0 is also equal to 0. Therefore, the product of these two terms approaches 0 as well.

Can this limit be evaluated using L'Hopital's rule?

Yes, this limit can be evaluated using L'Hopital's rule. By taking the derivative of both the numerator and denominator, and then evaluating the limit again, the result will still be 0.

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