Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1}$$

Short answer

$$ Answer: The value of the given limit is 2.

Step-by-step solution

Write the given limit

We are given the following limit: $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1}$$ We need to find the value of this limit or state that it does not exist.

Check for indeterminate form

Let's plug in x = 1 to see if we get an indeterminate form: $$\frac{\sqrt{10(1)-9}-1}{1-1} = \frac{\sqrt{1}-1}{0} = \frac{0}{0}$$ The given limit is in an indeterminate form (0/0), so we will need to perform algebraic manipulations to get rid of the indeterminate form before finding the limit.

Rationalize the numerator

We will rationalize the numerator by multiplying the top and bottom of the fraction with the conjugate of the numerator: $$\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1} \cdot \frac{\sqrt{10 x-9}+1}{\sqrt{10 x-9}+1}$$

Simplify the rationalized expression

Expand the numerator and denominator and simplify: On multiplying and expanding the expression we have, $$\lim _{x \rightarrow 1} \frac{(10 x - 9) - 2 \cdot \sqrt{10 x - 9} + 1}{(x - 1)(\sqrt{10 x - 9}+1)}$$ Simplify the numerator by combining the constant terms: $$\lim _{x \rightarrow 1} \frac{10x - 8 - 2\sqrt{10x - 9}}{(x - 1)(\sqrt{10x - 9} + 1)}$$

Factor out a common term from the numerator

To further simplify, we notice that there is a common factor in the numerator. Factor out 2 from the numerator: $$\lim _{x \rightarrow 1} \frac{2(5x - 4 - \sqrt{10x - 9})}{(x - 1)(\sqrt{10x - 9} + 1)}$$

Plug-in the limit value back into the expression

From Step 4 expression after simplification, we plug x = 1 back into the expression and see that we can now evaluate the limit: $$\lim _{x \rightarrow 1} \frac{2(5x - 4 - \sqrt{10x - 9})}{(x - 1)(\sqrt{10x - 9} + 1)} = \frac{2(5(1) - 4 - \sqrt{10(1) - 9})}{(1 - 1)(\sqrt{10(1) - 9} + 1)} = \frac{2(1)}{1} = 2$$ So, the limit is equal to 2: $$\lim _{x \rightarrow 1} \frac{\sqrt{10x - 9} - 1}{x - 1} = 2$$

Key concepts

Limit of a function

When exploring the concept of limits in calculus, we're looking at what happens to the value of a function as the input approaches a certain point. A limit is fundamental in calculus because it enables us to describe the behavior of functions at points where they may not be explicitly defined.

In the given exercise, \( \lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1} \), the aim is to determine the value the function approaches as x approaches 1. Direct substitution would yield an indeterminate form \( \frac{0}{0} \), which is undefined in the mathematical sense. To find the limit, we need to manipulate the function algebraically until we can substitute the value of x without getting an indeterminate form.

Understanding limits involves recognizing when to apply certain techniques such as factoring, expansion, or rationalizing to evaluate the limit effectively. In this example, the technique used is rationalization to simplify the expression and remove the indeterminate form to reveal the limit of the function as x approaches 1.

Indeterminate forms

Indeterminate forms occur when substituting a point into a function leads to an expression like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), which cannot be evaluated directly as they lack a clear value. In calculus, indeterminate forms prompt us to find different ways to manipulate the function to make it possible to evaluate the limit.

In our textbook exercise example, the direct substitution of x into \( \frac{\sqrt{10 x-9}-1}{x-1} \) resulted in the indeterminate form \( \frac{0}{0} \). Such forms often hint that the function is likely to simplify further, allowing us to find the limit. The ability to recognize and resolve indeterminate forms is essential in understanding limits and ensuring accurate calculation of the function's behavior around the point of interest.

Rationalizing the numerator

Rationalizing the numerator is a technique used to solve limits involving square roots, especially when facing an indeterminate form. It involves multiplying the numerator and denominator by the conjugate of the numerator to create a difference of squares, which often simplifies the expression.

For our problem, \( \lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1} \), rationalizing helps eliminate the square root in the numerator. After multiplying by the conjugate, we find that the problematic 'zero over zero' no longer exists and the expression simplifies nicely, allowing us to directly find the limit as x approaches 1.

This method is not just a trick for finding limits; it embodies the creative manipulation of algebraic expressions to reveal underlying relationships within a function near critical points. Rationalizing the numerator transforms the given function into a solvable form, reinforcing concepts of continuity and limits that are crucial in calculus.