Solve Limit with Square Root: \[\lim_{x\rightarrow -\infty }\sqrt{x^{2}+3}+x\]

In summary, a limit in calculus is used to determine the value that a function approaches at a specific point, rather than what it equals. To solve a limit with a square root, the expression must be simplified using algebraic manipulations. The limit may not always be defined if the expression inside the square root approaches a negative value. This type of limit differs from a regular limit in that the expression must be simplified before finding the limit. Solving a limit with a square root is significant in real-world applications as it allows us to analyze the behavior of a function and make informed decisions.
  • #1
Yankel
395
0
Hello

I am trying to solve this limit here:

\[\lim_{x\rightarrow -\infty }\sqrt{x^{2}+3}+x\]

I understand that it should be 0 since the power and square root cancel each other, while the power turned the minus into plus, and then when I add infinity I get 0. This is logic, I wish to know how to show it technically, if possible.

Thank you.
 
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  • #2
Multiply by $\frac{\sqrt{x^2+3}-x}{\sqrt{x^2+3}-x}$.
 
  • #3
Thank you, I didn't see it. :eek:
 

FAQ: Solve Limit with Square Root: \[\lim_{x\rightarrow -\infty }\sqrt{x^{2}+3}+x\]

What is a limit in calculus?

A limit in calculus is a fundamental concept that describes the behavior of a function as the input approaches a specific value. It is used to determine the value that a function approaches, rather than what it equals, at a particular point.

How do I solve a limit with a square root?

To solve a limit with a square root, we first need to simplify the expression using algebraic manipulations. In the given expression, we can factor out x from the square root, which will result in the limit being rewritten as: limx→-∞ x√(1+3/x2). Then, we can use the fact that limx→-∞ 1/x = 0 to simplify the expression further. Finally, we can substitute limx→-∞ x = -∞ to find the limit.

Is the limit of a square root always defined?

No, the limit of a square root may not always be defined. In order for the limit to exist, the expression inside the square root must approach a positive value as the input approaches the limit value. If the expression inside the square root approaches a negative value, the limit will not exist.

How is a limit with a square root different from a regular limit?

A limit with a square root is different from a regular limit in that the expression inside the square root must be simplified before finding the limit. This is because a square root cannot be evaluated for negative values, so the expression must be manipulated to ensure that the limit approaches a positive value.

What is the significance of solving a limit with a square root in real-world applications?

Solving a limit with a square root is useful in real-world applications because it allows us to analyze the behavior of a function as the input approaches a particular value. This can be used to predict the behavior of systems or processes and make informed decisions based on those predictions.

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