Question

Consider the function \(f(x)=\sqrt{x}\) (a) Is \(\lim _{x \rightarrow 0.25} \sqrt{x}=0.5\) a true statement? Explain. (b) Is \(\lim _{x \rightarrow 0} \sqrt{x}=0\) a true statement? Explain.

Short answer

Yes, both the statements are true. \(\lim _{x \rightarrow 0.25} \sqrt{x}=0.5\) and \(\lim _{x \rightarrow 0} \sqrt{x}=0\)

Step-by-step solution

Evaluate \(\lim _{x \rightarrow 0.25} \sqrt{x}\)

Plug in the value of x in the function \(\sqrt{x}\). This gives us \( \sqrt{0.25} = 0.5 \). It means as x approaches 0.25, the value of the function \( \sqrt{x} \) will become closer to 0.5.

Check if \(\lim _{x \rightarrow 0.25} \sqrt{x}=0.5\)

According to our evaluation from Step 1, as x approaches 0.25 from both sides, \( \sqrt{x} \) approaches 0.5, hence, it is true that \(\lim _{x \rightarrow 0.25} \sqrt{x}=0.5\).

Evaluate \(\lim _{x \rightarrow 0} \sqrt{x}\)

Similarly, plug the value of x in the function \( \sqrt{x} \). This gives us \( \sqrt{0} = 0 \). It means as x approaches 0, the value of \( \sqrt{x} \) will become closer to 0.

Check if \(\lim _{x \rightarrow 0} \sqrt{x}=0\)

According to our evaluation from Step 3, as x approaches 0 from both sides, \( \sqrt{x} \) approaches 0, hence, it is true that \(\lim _{x \rightarrow 0} \sqrt{x}=0\).

Key concepts

Limit of a Function

Understanding the limit of a function is crucial in calculus as it deals with the behavior of a function as the input approaches a particular point. It's like looking at what value a function is heading towards without necessarily reaching that point.

For example, in the exercise, we were asked to confirm if \(\lim _{x \rightarrow 0.25} \sqrt{x}=0.5\) and \(\lim _{x \rightarrow 0} \sqrt{x}=0\) were true statements. To do this, you plug in the approaching value into \(\sqrt{x}\), provided the function is defined at that point, to see what the output is. If the output corresponds to the value that \(x\) is approaching, you have found the limit.

It’s also important to note that limits can exist even if the function doesn't have a value at that point; what matters is the trend as \(x\) gets arbitrarily close, not the function's actual value at that point.

Square Root Function

The square root function, \(f(x) = \sqrt{x}\), is fundamental in calculus. This function assigns to each non-negative input \(x\) its non-negative square root. The domain of this function is all non-negative real numbers since square roots are not defined for negative numbers in the real number system.

The graph of a square root function forms a curve starting from the origin (0,0) and increases slowly at first, then more rapidly. It's key in many calculus problems due to its non-linear growth, which is distinct from polynomial functions.

Evaluating Limits

Evaluating limits is a process to determine the value that a function approaches as the input gets infinitely close to a certain number. There are several techniques to evaluate limits, including direct substitution, factoring, and rationalizing.

In our exercise, direct substitution was used. When we plugged \(0.25\) and \(0\) into the square root function, we got their respective roots \(0.5\) and \(0\), showing us the limits of the function at those points. For simple functions such as polynomials and root functions, direct substitution is often the clearest and quickest method. It's essential to be comfortable with these methods as they're the cornerstone of understanding more complex aspects of calculus.

Limits and Continuity

Limits and continuity are tightly knit concepts in the study of calculus. A function is continuous at a point if there is no abrupt change in its value. In other words, you should be able to draw the graph of the function at that point without lifting your pencil.

A key part of this is the limit. For a function to be continuous at a point \(x = a\), we need \(\lim _{x \rightarrow a} f(x) = f(a)\). This means that as \(x\) moves closer and closer to \(a\), \(f(x)\) approaches the value \(f(a)\), and \(f(a)\) is defined. From the exercise, we can see that the square root function is continuous at both \(0.25\) and \(0\) because the limits exist and match the function's value at these points.