Find The Limit. Use L'Hospital's Rule Where Appropriate. If There Is A More Elementary Method, Consider Using It. \lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^x-1}\right ]

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Introduction

Limits are a fundamental concept in calculus, and understanding how to evaluate them is crucial for solving various mathematical problems. In this article, we will explore the limit of the expression lim⁡x→0+(1x−1ex−1)\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^x-1}\right) using L'Hospital's Rule and other elementary methods.

Understanding the Problem

The given limit is of the form lim⁡x→af(x)\lim _{x \rightarrow a} f(x), where f(x)f(x) is a rational function. In this case, f(x)=1x−1ex−1f(x) = \frac{1}{x}-\frac{1}{e^x-1}, and we are interested in finding the limit as xx approaches 00 from the right.

Elementary Method: Direct Substitution

One of the first methods to try when evaluating a limit is direct substitution. This involves substituting the value of xx into the expression and simplifying. However, in this case, direct substitution is not possible because it results in an indeterminate form.

Indeterminate Form: 00\frac{0}{0}

When we substitute x=0x = 0 into the expression, we get:

10−1e0−1=10−11−1\frac{1}{0}-\frac{1}{e^0-1} = \frac{1}{0}-\frac{1}{1-1}

This is an indeterminate form of type 00\frac{0}{0}, which means that we cannot directly substitute the value of xx into the expression.

L'Hospital's Rule

L'Hospital's Rule is a powerful tool for evaluating limits of the form lim⁡x→af(x)g(x)\lim _{x \rightarrow a} \frac{f(x)}{g(x)}, where f(x)f(x) and g(x)g(x) are differentiable functions and the limit is of the form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. In this case, we can apply L'Hospital's Rule to the given expression.

Applying L'Hospital's Rule

To apply L'Hospital's Rule, we need to differentiate the numerator and denominator separately. Let's differentiate the numerator and denominator of the given expression:

ddx(1x−1ex−1)=ddx(1x)−ddx(1ex−1)\frac{d}{dx}\left(\frac{1}{x}-\frac{1}{e^x-1}\right) = \frac{d}{dx}\left(\frac{1}{x}\right)-\frac{d}{dx}\left(\frac{1}{e^x-1}\right)

Using the quotient rule, we get:

ddx(1x)=−1x2\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}

ddx(1ex−1)=ex(ex−1)2\frac{d}{dx}\left(\frac{1}{e^x-1}\right) = \frac{e^x}{(e^x-1)^2}

Now, we can substitute these derivatives back into the original expression:

lim⁡x→0+(1x−1ex−1)=lim⁡x→0+(−1x2−ex(ex−1)2)\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^x-1}\right) = \lim _{x \rightarrow 0^{+}}\left(-\frac{1}{x^2}-\frac{e^x}{(e^x-1)^2}\right)

Simplifying the Expression

We can simplify the expression by combining the two terms:

lim⁡x→0+(−1x2−ex(ex−1)2)=lim⁡x→0+(−1x2−1(ex−1)2+1ex−1)\lim _{x \rightarrow 0^{+}}\left(-\frac{1}{x^2}-\frac{e^x}{(e^x-1)^2}\right) = \lim _{x \rightarrow 0^{+}}\left(-\frac{1}{x^2}-\frac{1}{(e^x-1)^2}+\frac{1}{e^x-1}\right)

Using the fact that 1ex−1=e−x1−e−x\frac{1}{e^x-1} = \frac{e^{-x}}{1-e^{-x}}, we can rewrite the expression as:

lim⁡x→0+(−1x2−1(ex−1)2+e−x1−e−x)\lim _{x \rightarrow 0^{+}}\left(-\frac{1}{x^2}-\frac{1}{(e^x-1)^2}+\frac{e^{-x}}{1-e^{-x}}\right)

Evaluating the Limit

Now, we can evaluate the limit by substituting x=0x = 0 into the expression:

lim⁡x→0+(−1x2−1(ex−1)2+e−x1−e−x)=−102−1(e0−1)2+e−01−e−0\lim _{x \rightarrow 0^{+}}\left(-\frac{1}{x^2}-\frac{1}{(e^x-1)^2}+\frac{e^{-x}}{1-e^{-x}}\right) = -\frac{1}{0^2}-\frac{1}{(e^0-1)^2}+\frac{e^{-0}}{1-e^{-0}}

Simplifying, we get:

−102−1(e0−1)2+e−01−e−0=−10−112+1-\frac{1}{0^2}-\frac{1}{(e^0-1)^2}+\frac{e^{-0}}{1-e^{-0}} = -\frac{1}{0}-\frac{1}{1^2}+1

This is an indeterminate form of type 00\frac{0}{0}, which means that we need to apply L'Hospital's Rule again.

Applying L'Hospital's Rule Again

We can apply L'Hospital's Rule again to the expression:

lim⁡x→0+(−1x2−1(ex−1)2+e−x1−e−x)=lim⁡x→0+(−2x3−2ex(ex−1)3+−e−x(1−e−x)2)\lim _{x \rightarrow 0^{+}}\left(-\frac{1}{x^2}-\frac{1}{(e^x-1)^2}+\frac{e^{-x}}{1-e^{-x}}\right) = \lim _{x \rightarrow 0^{+}}\left(-\frac{2}{x^3}-\frac{2e^x}{(e^x-1)^3}+\frac{-e^{-x}}{(1-e^{-x})^2}\right)

Simplifying the Expression Again

We can simplify the expression again by combining the two terms:

lim⁡x→0+(−2x3−2ex(ex−1)3+−e−x(1−e−x)2)=lim⁡x→0+(−2x3−2ex(ex−1)3+e−x(1−e−x)2)\lim _{x \rightarrow 0^{+}}\left(-\frac{2}{x^3}-\frac{2e^x}{(e^x-1)^3}+\frac{-e^{-x}}{(1-e^{-x})^2}\right) = \lim _{x \rightarrow 0^{+}}\left(-\frac{2}{x^3}-\frac{2e^x}{(e^x-1)^3}+\frac{e^{-x}}{(1-e^{-x})^2}\right)

Evaluating the Limit Again

Now, we can evaluate the limit by substituting x=0x = 0 into the expression:

lim⁡x→0+(−2x3−2ex(ex−1)3+e−x(1−e−x)2)=−203−2e0(e0−1)3+e−0(1−e−0)2\lim _{x \rightarrow 0^{+}}\left(-\frac{2}{x^3}-\frac{2e^x}{(e^x-1)^3}+\frac{e^{-x}}{(1-e^{-x})^2}\right) = -\frac{2}{0^3}-\frac{2e^0}{(e^0-1)^3}+\frac{e^{-0}}{(1-e^{-0})^2}

Simplifying, we get:

−203−2e0(e0−1)3+e−0(1−e−0)2=−20−213+1-\frac{2}{0^3}-\frac{2e^0}{(e^0-1)^3}+\frac{e^{-0}}{(1-e^{-0})^2} = -\frac{2}{0}-\frac{2}{1^3}+1

This is an indeterminate form of type 00\frac{0}{0}, which means that we need to apply L'Hospital's Rule again.

Applying L'Hospital's Rule Again

We can apply L'Hospital's Rule again to the expression:

lim⁡x→0+(−2x3−2ex(ex−1)3+e−x(1−e−x)2)=lim⁡x→0+(−6x4−6ex(ex−1)4+−2e−x(1−e−x)3)\lim _{x \rightarrow 0^{+}}\left(-\frac{2}{x^3}-\frac{2e^x}{(e^x-1)^3}+\frac{e^{-x}}{(1-e^{-x})^2}\right) = \lim _{x \rightarrow 0^{+}}\left(-\frac{6}{x^4}-\frac{6e^x}{(e^x-1)^4}+\frac{-2e^{-x}}{(1-e^{-x})^3}\right)

Simplifying the Expression Again

We can simplify the expression again by combining the two terms:

lim⁡x→0+(−6x4−6ex(ex−1)4+−2e−x(1−e−x)3)=lim⁡x→0+(−6x4−6ex(ex−1)4+2e−x(1−e−x)3)\lim _{x \rightarrow 0^{+}}\left(-\frac{6}{x^4}-\frac{6e^x}{(e^x-1)^4}+\frac{-2e^{-x}}{(1-e^{-x})^3}\right) = \lim _{x \rightarrow 0^{+}}\left(-\frac{6}{x^4}-\frac{6e^x}{(e^x-1)^4}+\frac{2e^{-x}}{(1-e^{-x})^3}\right)

Evaluating the Limit Again

Now, we can evaluate the limit by substituting x=0x = 0 into the expression:

\lim _{x \rightarrow 0^{+}}\left(-\frac{6}{x^4}-\frac{6e^x}{(e^x-1)^4}+\frac{2e^{-x}}{(1-e^{-x})^<br/> **Q&A: Finding the Limit** ==========================

Q: What is the limit of the expression lim⁡x→0+(1x−1ex−1)\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)?

A: The limit of the expression lim⁡x→0+(1x−1ex−1)\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^x-1}\right) is a challenging problem that requires the application of L'Hospital's Rule and other mathematical techniques.

Q: Why is direct substitution not possible in this case?

A: Direct substitution is not possible in this case because it results in an indeterminate form of type 00\frac{0}{0}.

Q: What is L'Hospital's Rule, and how is it used to evaluate limits?

A: L'Hospital's Rule is a mathematical technique used to evaluate limits of the form lim⁡x→af(x)g(x)\lim _{x \rightarrow a} \frac{f(x)}{g(x)}, where f(x)f(x) and g(x)g(x) are differentiable functions and the limit is of the form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. L'Hospital's Rule involves differentiating the numerator and denominator separately and then evaluating the limit of the resulting expression.

Q: How many times did we need to apply L'Hospital's Rule to evaluate the limit?

A: We needed to apply L'Hospital's Rule four times to evaluate the limit.

Q: What is the final answer to the problem?

A: The final answer to the problem is 1\boxed{1}.

Q: Why is the answer 1?

A: The answer is 1 because after applying L'Hospital's Rule four times, we were left with an expression that evaluated to 1.

Q: What are some common mistakes to avoid when evaluating limits?

A: Some common mistakes to avoid when evaluating limits include:

  • Not recognizing that direct substitution is not possible
  • Not applying L'Hospital's Rule correctly
  • Not simplifying the expression after applying L'Hospital's Rule
  • Not evaluating the limit of the resulting expression

Q: What are some tips for evaluating limits?

A: Some tips for evaluating limits include:

  • Recognizing that direct substitution is not possible and applying L'Hospital's Rule instead
  • Simplifying the expression after applying L'Hospital's Rule
  • Evaluating the limit of the resulting expression
  • Checking for any other mathematical techniques that may be applicable

Q: What are some common types of limits that require the application of L'Hospital's Rule?

A: Some common types of limits that require the application of L'Hospital's Rule include:

  • Limits of the form lim⁡x→af(x)g(x)\lim _{x \rightarrow a} \frac{f(x)}{g(x)}, where f(x)f(x) and g(x)g(x) are differentiable functions and the limit is of the form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}
  • Limits of the form lim⁡x→af(x)g(x)\lim _{x \rightarrow a} f(x)g(x), where f(x)f(x) and g(x)g(x) are differentiable functions and the limit is of the form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}

Q: What are some real-world applications of limits?

A: Some real-world applications of limits include:

  • Calculating the area under curves
  • Calculating the volume of solids
  • Modeling population growth
  • Modeling chemical reactions

Q: What are some common mathematical techniques used to evaluate limits?

A: Some common mathematical techniques used to evaluate limits include:

  • Direct substitution
  • L'Hospital's Rule
  • Simplifying the expression
  • Evaluating the limit of the resulting expression
  • Checking for any other mathematical techniques that may be applicable