Question

Verify that the assumptions of L'Hôpital's rule hold, and find the following limits. (a) \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (e-x)+x-1} ;\) (b) \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}\) (c) \(\lim _{x \rightarrow 0} \frac{(1+x)^{1 / x}-e}{x} ;\) (d) \(\lim _{x \rightarrow 0^{+}}\left(x^{q} \ln x\right), q>0\) (e) \(\lim _{x \rightarrow+\infty}\left(x^{-q} \ln x\right), q>0\) (f) \(\lim _{x \rightarrow 0^{+}} x^{x}\) (g) \(\lim _{x \rightarrow+\infty}\left(x^{q} a^{-x}\right), a>1, q>0\) (h) \(\lim _{x \rightarrow 0}\left(\frac{1}{x^{2}}-\operatorname{cotan}^{2} x\right)\); (i) \(\lim _{x \rightarrow+\infty}\left(\frac{\pi}{2}-\arctan x\right)^{1 / \ln x}\) (j) \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{1 /(1-\cos x)}\).

Short answer

\(\frac{2e}{e-1}\) using L'Hôpital's Rule for (a).

Step-by-step solution

Identify indeterminate form

We begin with exercise (a): \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (e-x)+x-1}\). First, substitute \(x = 0\) into the function: \(\frac{e^{0}-e^{0}}{\ln (e-0)+0-1} = \frac{0}{0}\). This is an indeterminate form, so L'Hôpital's Rule can be applied.

Verify applicability of L'Hôpital's rule

L'Hôpital's rule requires that both the numerator and denominator are differentiable near \(x = 0\) and that the limit we are trying to find results in an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). In this case, both \(e^x-e^{-x}\) and \(\ln(e-x)+x-1\) are differentiable near \(x=0\), satisfying conditions for L'Hôpital's rule.

Differentiate numerator and denominator

Differentiate the numerator: \(\frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x}\). Differentiate the denominator: \(\frac{d}{dx}(\ln(e-x) + x - 1) = -\frac{1}{e-x} + 1\). Thus, the derivative of the denominator is \(1 - \frac{1}{e-x}\).

Apply L'Hôpital's Rule and evaluate

Now apply L'Hôpital's Rule: \(\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}}{1-\frac{1}{e-x}}.\)Substitute \(x=0\) into this expression:\(\lim _{x \rightarrow 0} \frac{e^0 + e^0}{1 - \frac{1}{e-0}} = \frac{2}{1 - \frac{1}{e}}. \)Simplify: \(\frac{2e}{e-1}\). Hence, the limit is \(\frac{2e}{e-1}\).

Conclusion

The limit for the given expression \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (e-x)+x-1}\) is verified using L'Hôpital's Rule, resulting in the value \(\frac{2e}{e-1}\).

Key concepts

Indeterminate Forms

Indeterminate forms are crucial in understanding the need for tools like L'Hôpital's Rule in calculus. They occur when evaluating certain limits that initially don't give a clear numerical value, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms are indeterminate because they can result in any number at all, depending on how the expressions in the limit behave as they approach the point of interest.
Common indeterminate forms include:

• \(\frac{0}{0}\) - arising when both the numerator and denominator tend to zero.
• \(\frac{\infty}{\infty}\) - appearing when both approach infinity.
• \(0^0\) - especially in exponential contexts.
• \(\infty - \infty\) - when two infinite quantities subtract each other.
• \(1^\infty\) - seen in growth situations like compound interest.
• \(0 \times \infty\) - where a zero quantity multiplies an infinite one.

For instance, in exercise (a) \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (e-x)+x-1}\), substituting \(x = 0\) yields \(\frac{0}{0}\), confirming an indeterminate form. This is where L'Hôpital's Rule becomes useful, allowing us to differentiate the numerator and denominator separately to evaluate the limit.

Limits in Calculus

Limits in calculus are foundational, allowing us to analyze the behavior of functions as they approach a specific point. They help in understanding the notion of continuity, derivatives, and integrals. At its core, a limit investigates the value a function \(f(x)\) approaches as \(x\) gets closer to a particular number.
Key ideas in limits include:

• Existence of a Limit: For \(\lim_{x \to c}f(x)\) to exist, \(f(x)\) must approach the same value from both the left and the right of \(c\).
• One-Sided Limits: These consider the behavior of \(f(x)\) as \(x\) approaches \(c\) from just one side - either the left (\(x \to c^-\)) or the right (\(x \to c^+\)).
• Infinite Limits: A limit can be infinite if \(f(x)\) or \(g(x)\) grows without bound as \(x\) approaches \(c\).

In our problem, applying limits helps determine the expression's behavior near potentially undefined points. In L'Hôpital's Rule, limits turn an indeterminate form into a more tractable one through differentiation, refining the analysis of such functions.

Differentiable Functions

A differentiable function is one that has a derivative at every point in its domain, meaning it can be smoothly graphed without any breaks, jumps, or sharp corners. Differentiability implies the function is also continuous, ensuring there are no abrupt changes in direction.
Important concepts include:

• Derivative: The derivative represents the instantaneous rate of change of a function at a given point, typically denoted as \(f'(x)\) or \(\frac{df}{dx}\).
• Smoothness: A differentiable function's graph will not have sharp edges, and its tangent line exists at every point.
• Continuity: If \(f(x)\) is differentiable at \(x = c\), it must be continuous at \(c\), though the converse may not hold.

L'Hôpital's Rule requires both the numerator and denominator of an expression to be differentiable when seeking to solve an indeterminate limit. For example, when differentiating \(e^x - e^{-x}\) and \(\ln(e-x) + x - 1\), confirming their differentiability ensures the correct application of the rule. This verification step is critical in determining the result of the limit accurately.