Question

Limits Evaluate the following limits. Use l'Hópital's Rule when it is comvenient and applicable. $$\lim _{x \rightarrow 1^{+}}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right)$$

Short answer

Question: Evaluate the limit as x approaches 1 from the positive side: \(\lim _{x \rightarrow 1^{+}}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right)\). Answer: The limit does not exist.

Step-by-step solution

Combine the two fractions into a single fraction

To combine the fractions, find a common denominator which is \((x-1)\sqrt{x-1}\). Then, the expression becomes: $$ \frac{1}{x-1}-\frac{1}{\sqrt{x-1}} = \frac{\sqrt{x-1} - (x-1)}{(x-1)\sqrt{x-1}} $$

Simplify the numerator

Now, multiply the numerator and denominator by the conjugate of the numerator to get a difference of squares: $$ \frac{\sqrt{x-1} - (x-1)}{(x-1)\sqrt{x-1}} \cdot \frac{\sqrt{x-1} + (x-1)}{\sqrt{x-1}+(x-1)} = \frac{(x-1)^2 - (\sqrt{x-1})^2}{(x-1)^2\sqrt{x-1}} $$ Simplifying the expression: $$ \frac{x^2 - 2x + 1 - (x-1)}{(x-1)^2\sqrt{x-1}} = \frac{x^2 - 3x + 2}{(x-1)^2\sqrt{x-1}} $$

Factor the numerator

Factor the numerator to find: $$ \frac{(x-1)(x-2)}{(x-1)^2\sqrt{x-1}} $$

Simplify the expression

Simplify by canceling out the common terms: $$ \frac{x-2}{(x-1)\sqrt{x-1}} $$

Evaluate the limit

Now we can find the limit as x approaches 1 from the positive side: $$ \lim _{x \rightarrow 1^{+}}\left(\frac{x-2}{(x-1)\sqrt{x-1}}\right) = \frac{1-2}{(1-1)\sqrt{1-1}} = \frac{-1}{0} $$ Since dividing by zero is undefined, the limit does not exist.

Key concepts

l'Hôpital's Rule

Understanding l'Hôpital's Rule is crucial for solving limits that initially seem indeterminate, often appearing as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule states that for limits of the form \( \lim_{x \to c} \frac{f(x)}{g(x)} \) when both \( f(c) = 0 \) and \( g(c) = 0 \) or both approach infinity, we can compute \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), where \( f'(x) \) and \( g'(x) \) are derivatives of \( f \) and \( g \), respectively.
This method is handy when standard algebraic simplifications are not enough. However, remember to first check that the conditions for applying l'Hôpital's Rule are met. In many exercises, combining other techniques like simplification or factoring can aid before applying l'Hôpital's Rule if necessary.
In the given exercise, the limit evaluates to an indeterminate form, but the process of simplifying the expression solves it without needing the rule. It is always a good idea to simplify first, as it might make applying or forgoing l'Hôpital's Rule clear.

Fraction Simplification

Fraction simplification involves reducing a complex fraction into a simpler form. This process includes finding a common denominator, factorizing expressions, and canceling out terms when possible.
For example, in the original exercise, the two separate fractions \( \frac{1}{x-1} \) and \( \frac{1}{\sqrt{x-1}} \) are combined using a common denominator \((x-1)\sqrt{x-1}\). This step is pivotal as it allows the expression to be rewritten in a single fraction, making it more manageable.
The numerator \( \sqrt{x-1} - (x-1) \) should then be simplified further, often by multiplying by a conjugate, which gets us closer to a form where limit evaluation is straightforward.
It's crucial to carefully perform each algebraic manipulation step-by-step to avoid errors, ensuring you correctly simplify to gain truly equivalent expressions.

Common Denominator

Finding a common denominator is a foundational skill in manipulating fractions. It involves identifying a shared factor in the denominators of multiple fractions to combine them into a single expression.
In our exercise, identifying \((x-1) \sqrt{x-1}\) as the common denominator allowed us to combine the fractions into a single, more straightforward form. This process enables us to manipulate and simplify the resulting expression more effectively.
Once fractions share a common denominator, their numerators can be straightforwardly combined, allowing us to further simplify or solve the problem at hand. Ensure that every term in your expression shares this common denominator to avoid discrepancies.

Difference of Squares

Using the difference of squares is a straightforward technique often applied in algebra to simplify expressions. The formula \(a^2 - b^2 = (a-b)(a+b)\) helps in breaking down complex expressions.
In the exercise's solution, multiplying the numerator by its conjugate \(\sqrt{x-1} + (x-1)\) leads to creating a difference of squares which simplifies the numerator to \((x-1)^2 - (\sqrt{x-1})^2\).
This simplification step is powerful, especially when dealing with limits, as it often reveals opportunities to cancel terms that make the initial expression resolve into a form ready for limit evaluation.
Applying the difference of squares is not only an effective tool for solving exercises but also underlines the importance of recognizing patterns within expressions – a crucial skill in simplifying mathematical problems.