Question

Evaluate the following limits. \(\lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1}\)

Short answer

Answer: The limit of the function is 2.

Step-by-step solution

Identify the limit to evaluate

The problem requires us to evaluate the following limit: \(\lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1}\)

Attempt direct substitution

If we attempt to substitute the limit point, \(x=1\), directly into the function, we get: \(\frac{1-1}{\sqrt{1}-1} = \frac{0}{0}\) This is an indeterminate form, meaning that we need to manipulate the expression to simplify it or use other methods to find the limit.

Rationalize the denominator

Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{x}+1\), in order to eliminate the radical from the denominator: \(\lim_{x \rightarrow 1} \frac{(x-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

Simplify

Now we can use the distributive property and difference of squares to simplify the expression: \(\lim_{x \rightarrow 1} \frac{(x-1)(\sqrt{x}+1)}{x-1}\) Notice that we get cancelation of terms in the numerator and denominator: \(\lim_{x \rightarrow 1} \sqrt{x}+1\)

Evaluate the limit

We can now directly substitute the limit point, \(x=1\), into the simplified expression to find the limit: \(\lim_{x \rightarrow 1} \sqrt{x}+1 = \sqrt{1}+1 = 1+1 = 2\) Therefore, the limit of the function is 2: \(\lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1} = 2\)

Key concepts

Limits and Continuity

Understanding the concept of limits is a foundational aspect of calculus. A limit evaluates what value a function approaches as the input approaches a particular point. In many cases, we can directly substitute the value into the function to find the limit. However, when direct substitution results in an undefined value, such as divisibility by zero, other techniques must be employed.

Continuity, on the other hand, means a function is unbroken or uninterrupted at a point, and we can say a function is continuous at a point if the limit matches the function's value at that point. For example, a function is continuous at a point 'a' if \( \lim_{x \rightarrow a} f(x) = f(a) \). In the given problem, understanding the concept of continuity helps identify that the function is not continuous at \(x=1\) because it leads to an indeterminate form, prompting further analysis.

Indeterminate Forms

Indeterminate forms occur when limits produce an undefined or ambiguous result. The most common types are \(0/0\) and \(\infty/\infty\), but there are several others, such as \(0 \times \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\). These forms do not give insight into the behavior of limits, and they indicate that further simplification or application of special techniques is necessary to find a limit's value.

In the given exercise, we encounter the \(0/0\) indeterminate form upon direct substitution. This suggests that we need to manipulate the expression to successfully evaluate the limit, signaling that the function's behavior near \(x = 1\) is more complex than the initial form suggests.

Rationalization

Rationalization is a technique used to eliminate radicals from the denominator of a fraction. The methodology involves multiplying both the numerator and denominator by the 'conjugate' of the denominator. In our exercise, the denominator \(\sqrt{x}-1\) is a radical expression, and we multiply by its conjugate \(\sqrt{x}+1\) to rationalize it.

This process takes advantage of the difference of squares identity \(a^2 - b^2 = (a + b)(a - b)\), which results in a rational expression — one that does not have a square root in the denominator. After rationalization, we can often cancel terms and simplify the expression, making it possible to directly substitute the value of the limit. It's an essential algebraic tool in limit evaluation.

Algebra of Limits

The algebra of limits refers to a set of rules and techniques used to manipulate and solve limits algebraically. This includes rules for sum, difference, product, and quotient of limits, as well as the power and root of limits. For instance, if the limits of functions \(f(x)\) and \(g(x)\) as \(x\) approaches the same value are known, we can find the limit of \(f(x) \pm g(x)\), \(f(x)g(x)\), and \(\frac{f(x)}{g(x)}\), provided that \(g(x)\) is not zero at that point.

These rules help us in solving complex limit problems by breaking them down into simpler parts that are easier to manage. During the rationalization process in our example, the algebra of limits is implicitly used to combine and cancel terms, ultimately leading to a solution that can be directly evaluated.