Question

Evaluating limits Evaluate the following limits, where \(c\) and \(k\) are constants. $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1}$$

Short answer

Question: Evaluate the limit $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1}$$ Answer: The limit is equal to 5.

Step-by-step solution

Identify the problem

We want to evaluate the limit: $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1}$$ Direct substitution of \(x=1\) in this expression results in an indeterminate form (0/0).

Rationalize the Numerator

To fix the indeterminate form, we will multiply the expression by the conjugate of the numerator. The conjugate of \(\sqrt{10x-9}-1\) is \(\sqrt{10x-9}+1\). So we multiply both the numerator and denominator by this conjugate: $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1} \times \frac{\sqrt{10x-9}+1}{\sqrt{10x-9}+1}$$

Simplify the expression

After multiplying, simplify the resulting expression by expanding and simplifying the numerator and denominator: $$\lim _{x \rightarrow 1} \frac{(10x-9)-1(\sqrt{10x-9}+1)(\sqrt{10x-9}+1)}{(x-1)(\sqrt{10x-9}+1)}$$ Now, after cancelling out the terms in the numerator we get: $$\lim _{x \rightarrow 1} \frac{10x-10}{(x-1)(\sqrt{10x-9}+1)}$$

Factor out a 10 in the numerator

Factoring out 10 in the numerator gives the following expression: $$\lim _{x \rightarrow 1} \frac{10(x-1)}{(x-1)(\sqrt{10x-9}+1)}$$

Cancel out the common term

Now we can cancel the common term \((x-1)\) from the numerator and denominator: $$\lim _{x \rightarrow 1} \frac{10}{\sqrt{10x-9}+1}$$

Evaluate the limit

Now, we can substitute \(x=1\) into the simplified expression: $$\lim _{x \rightarrow 1} \frac{10}{\sqrt{10(1)-9}+1} = \frac{10}{\sqrt{1}+1} = \frac{10}{2}$$ Finally, we get the value of the limit as: $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1} = 5$$

Key concepts

Indeterminate Forms

When dealing with limits in calculus, it’s common to encounter what are known as indeterminate forms. These occur when substituting a value into a function results in an expression like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Such expressions don’t give you a clear number, making it tricky to determine the limit directly.

Imagine trying to compute \( \lim_{x \to 1} \frac{\sqrt{10x-9}-1}{x-1} \). By substituting \( x = 1 \), you get \( \frac{0}{0} \), indicating an indeterminate form. These forms don't show us much without further manipulation of the function, often requiring different techniques to resolve.

Rationalization

Rationalization is a technique used to simplify complex expressions, frequently to resolve indeterminate forms. In this context, it’s all about getting rid of square roots or other irrational components in fractions.

For instance, if you face a limit problem like \( \frac{\sqrt{10x-9}-1}{x-1} \), you can use the conjugate of the numerator to eliminate the square root. The conjugate of \( \sqrt{10x-9}-1 \) is \( \sqrt{10x-9}+1 \), and multiplying both the numerator and the denominator of the expression by this conjugate rationalizes the numerator.

This multiplication converts tricky square-root terms into a more manageable form without changing the limit you're trying to evaluate, as it effectively transforms them into a polynomial form which is easier to work with.

Limit Evaluation Method

The techniques of limit evaluation often involve simplifying expressions until they no longer result in indeterminacy. Once you've rationalized the expression, like in our example, the next step is simplification.

After multiplying through with the conjugate, you end up with something like \( \frac{10(x-1)}{(x-1)(\sqrt{10x-9}+1)} \). Carefully factor out common elements, such as \( x-1 \), simplifying the expression significantly.

The following step involves re-evaluation, substituting the limiting value, \( x=1 \), into the simplified function: \( \lim_{x \to 1} \frac{10}{\sqrt{10x-9}+1} \). Because it's now simplified, the function can safely accept the substitution, yielding a finite result like 5.

These methods combine to rigorously resolve limits, transforming indeterminate forms into real numeric answers using algebraic manipulation and substitution.