Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 0} \frac{x}{\sqrt{c x+1}-1}, c \neq 0$$

Short answer

Answer: The limit of the given function as x approaches 0 is \(\frac{2}{c}\).

Step-by-step solution

Identify the indeterminate form

The given function is: $$\lim_{x \rightarrow 0} \frac{x}{\sqrt{c x+1}-1}$$ As x approaches 0, the function has the indeterminate form \(\frac{0}{0}\). To evaluate this limit, we will use a technique called rationalizing the denominator.

Rationalize the denominator

Multiply the fraction by its conjugate, which is \(\frac{\sqrt{c x+1}+1}{\sqrt{c x+1}+1}\): $$\lim_{x \rightarrow 0} \frac{x}{\sqrt{c x+1}-1} \cdot \frac{\sqrt{c x+1}+1}{\sqrt{c x+1}+1}$$

Simplify the expression

Expand the numerator and denominator: $$\lim_{x \rightarrow 0} \frac{x(\sqrt{c x+1}+1)}{(c x+1)-1}$$ Now, simplify the denominator: $$\lim_{x \rightarrow 0} \frac{x(\sqrt{c x+1}+1)}{c x}$$

Cancel the common factors

We can cancel the common factor x in both the numerator and the denominator: $$\lim_{x \rightarrow 0} \frac{\sqrt{c x+1}+1}{c}$$

Evaluate the limit

Now that the expression is simplified, we can find the limit by directly substituting x = 0: $$\lim_{x \rightarrow 0} \frac{\sqrt{c x+1}+1}{c} = \frac{\sqrt{c(0)+1}+1}{c} = \frac{\sqrt{1}+1}{c} = \frac{2}{c}$$ Thus, the limit of the given function as x approaches 0 is: $$\lim_{x \rightarrow 0} \frac{x}{\sqrt{c x+1}-1} = \frac{2}{c}$$

Key concepts

Indeterminate Form

When calculating limits in calculus, you might encounter an indeterminate form, which is a mathematical expression that is not strictly defined until further analysis is done. A familiar indeterminate form is \(\frac{0}{0}\). This arises when both the numerator and the denominator of a fraction trend towards zero as the variable approaches a certain point. In our exercise, as x gets closer to zero, \(\lim_{x \rightarrow 0} \frac{x}{\sqrt{c x+1}-1}\) becomes a \(\frac{0}{0}\) situation, which can't be evaluated just by substitution.

To resolve the indeterminacy and find the limit, we perform operations that transform the original expression into an equivalent one that is no longer indeterminate. This often involves algebraic manipulation like factoring, expanding, and in many cases, rationalizing the denominator, which is illustrated in the step-by-step solution provided. By addressing the indeterminate form effectively, it becomes possible to evaluate the limit.

Rationalizing the Denominator

The process of rationalizing the denominator involves removing the square root (or cube root, etc.) from the bottom of a fraction. This is achieved by multiplying the fraction by a form of 1 that contains the conjugate of the denominator. The conjugate of \(\sqrt{c x+1}-1\) is \(\sqrt{c x+1}+1\). By doing this in our exercise, we eliminate the square root from the denominator, transforming our indeterminate form into one that we can work with more easily. It often simplifies the expression, resulting in an identical yet simpler version, which can then be further manipulated towards finding the limit. The conjugate not only clears the square root but also sets the stage for canceling out common factors between the numerator and the denominator, which is a crucial step in limit evaluation.

Limit of a Function

The limit of a function is a fundamental concept in calculus that describes the behavior of a function as its argument approaches a certain point, but does not necessarily reach that point. This concept is especially useful for studying functions at points where they are not well-defined, such as points of discontinuity or, as in our case, indeterminate forms. Limits can often be found by direct substitution if they are not indeterminate. However, when direct substitution leads to an indeterminate form, additional techniques – like the aforementioned rationalization – are required. Once the limit is properly expressed without indeterminate forms, it can then be evaluated with ease. In the context of our exercise, the limit of the function as x approaches 0 was ultimately simplified and evaluated by substitution.

Algebraic Simplification

The process of algebraic simplification is crucial in making complex expressions more accessible and easier to evaluate. In limit evaluation, simplification might involve expanding products, factoring polynomials, and canceling out alike terms. In our exercise, after rationalizing the denominator, the expression is further simplified by expanding and then canceling the common factor of x from both the numerator and the denominator. This reduces the expression to a much simpler form that is no longer indeterminate, thus allowing for the limit to be evaluated successfully. It's important to proceed with these simplifications carefully, ensuring each step is mathematically valid to avoid any errors that could lead to incorrect conclusions. The ultimate goal is to reach a point where the limit can be evaluated directly and accurately.