91Ó°ÊÓ

The square root function, often represented as \( \sqrt{y} \), is a mathematical operation that finds a number which, when multiplied by itself, yields the original number \( y \). For real numbers, this function is usually defined for non-negative numbers. This is because the square root of a negative number is not real. Hence, we focus on \( y \geq 0 \), where the square root is continuous and increasing.

• For \( y = 0 \), \( \sqrt{y} = 0 \).
• For \( y > 0 \), \( \sqrt{y} > 0 \).
• The function grows slowly for larger values of \( y \).

This means as we have inputs approaching zero from the right (i.e., very small positive numbers), the output of the square root will come close to zero. This behavior will assist in finding the limit when involving expressions under square roots, such as in problems encountered in calculus.

Continuity in calculus is a property of functions that intuitively means no breaks, holes, or jumps. A function is continuous at a point \( x = a \) if the following holds:

• The function \( f(x) \) is defined at \( x = a \).
• The limit of \( f(x) \) as \( x \) approaches \( a \) exists.
• The limit and the function value agree, i.e., \( \lim_{x \to a}f(x) = f(a) \).

A continuous function is nice to work with because you can substitute values directly into the function to find limits, like how we approached \( \sqrt{\frac{x-1}{x+2}} \) as \( x \to 1^+ \). The continuous nature of the square root ensures that as the expression inside approaches zero from the right, the square root itself approaches zero smoothly without any disruption.

A right-hand limit is concerned with what happens to a function as the variable approaches a particular value from the right, indicated by the notation \( x \to a^+ \). This essentially means we look at values of \( x \) that are slightly greater than \( a \).

• For \( x \to 1^+ \), we only consider \( x \) values that are more than, but close to, 1.
• It provides insights into the behavior of the function as it nears the point from one direction.

In the given exercise, \( \lim_{x \to 1^+} \sqrt{\frac{x-1}{x+2}} \), the focus was on \( x \) values greater than 1. Here, since \( x - 1 > 0 \), eventually, the expression \( \frac{x - 1}{x + 2} \) approaches \( 0^+ \). This positive infinitesimal argument under the square root implies the function remains smooth and predictable, guiding to the conclusion that the limit is zero.