Problem 62 - Calculus Early Transcendentals

Question

Evaluate the following limits or explain why they do not exist. Check your results by graphing. $$\lim _{x \rightarrow 0}\left(e^{5 x}+x\right)^{1 / x}$$

Step-by-Step Solution

Answer

Question: Evaluate the limit: $$\lim _{x \rightarrow 0}\left(e^{5 x}+x\right)^{1 / x}$$ Answer: The limit as x approaches 0 is equal to 1/e.

Step 1: Use Property of Logarithms

We know that, if y = a^x, then x = log_a(y). This will help us to change the exponent part to logarithm which will further help us to solve the limit. $$\lim _{x \rightarrow 0}\left(e^{5 x}+x\right)^{1 / x} =\lim _{x \rightarrow 0}e^{(\frac{1}{x}\log (e^{5 x}+x))}$$ We've transformed the limit into an exponential function, hence, if we find its exponent limit, we could easily find the value of the expression as well.

Step 2: Simplifying the new limit

Now, our goal is to find the limit in the exponent: $$\lim _{x \rightarrow 0}\frac{1}{x} \log \left(e^{5 x}+x\right)$$ Using the definition of the natural logarithm, we have: $$\lim _{x \rightarrow 0}\frac{1}{x}\log \left(1+e^{5x}x\right)$$ Now, we still have an indeterminate form 0 times infinity. Therefore, let's rewrite the fraction using the reciprocal of x.

Step 3: Rewrite the fraction

Rewrite the limit of the exponent by using the reciprocal of x. This allows us to apply L'Hopital's rule: $$\lim _{x \rightarrow 0}\frac{\log \left(1+e^{5x}x\right)}{\frac{1}{x}}$$

Step 4: Apply L'Hopital's Rule

Apply L'Hopital's rule by taking the derivative of the numerator and denominator: $$\lim _{x \rightarrow 0}\frac{\frac{d}{dx} \log \left(1+e^{5x}x\right)}{\frac{d}{dx} \frac{1}{x}}$$ The derivative of the numerator is: $$\frac{1}{1+e^{5x}x}\cdot\left(5e^{5x}x+e^{5x}\right)$$ The derivative of the denominator is: $$-\frac{1}{x^2}$$ Now, substitute the expressions into the limit.

Step 5: Simplify the limit

After substituting the derivatives of the numerator and denominator into the limit, we now evaluate and simplify the limit: $$\lim _{x \rightarrow 0}\frac{\frac{1}{1+e^{5x}x}\cdot\left(5e^{5x}x+e^{5x}\right)}{-\frac{1}{x^2}}$$ Cancel out the x from the denominator and the x on the second term of the numerator: $$\lim _{x \rightarrow 0}\frac{\frac{1}{1+e^{5x}x}\cdot\left(5e^{5x}x^2+e^{5x}x\right)}{-x}$$ Now we can evaluate the limit: $$\left.\frac{\frac{1}{1+e^{5x}x}\cdot\left(5e^{5x}x^2+e^{5x}x\right)}{-x}\right|_{x=0} =-\frac{1}{1} = -1$$ The limit of the exponent is -1, now, put the exponent back to the original equation.

Step 6: Final answer

Finally, put the exponent back in its place and simplify to get the final result: $$\lim _{x \rightarrow 0}\left(e^{5 x}+x\right)^{1 / x} = e^{-1} = \frac{1}{e}$$ So the limit as x approaches 0 is equal to 1/e.

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is an essential tool in calculus for evaluating limits that result in indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule suggests that under certain conditions, the limit of a quotient of two functions can be found by differentiating the numerator and the denominator separately, and then taking the limit of their quotient. However, it's crucial to remember:

• L'Hopital's Rule only applies if the limit initially results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• The derivatives must exist near the point of interest.
• Reapplication of the rule might be necessary if indeterminate forms persist after differentiation.

In the given exercise, we have applied L'Hopital's Rule to evaluate the limit \( \lim _{x \rightarrow 0}\frac{\log \left(1+e^{5x}x\right)}{\frac{1}{x}} \). By differentiating the numerator and denominator, as shown, we can simplify the expression significantly, leading to a clear evaluation of the limit.

Natural Logarithm

The natural logarithm, denoted as \(\log\) or \(\ln\), is the logarithm to the base \(e\), where \(e\) is approximately 2.71828. It is widely used in mathematics due to its unique properties, especially in calculus and mathematical analysis.

• It helps in transforming exponential relationships into linear ones, making them easier to handle mathematically.
• The derivative of \(\ln(x)\) is \(\frac{1}{x}\), which is a fundamental concept in calculus.

In this problem, taking the natural logarithm of \( \left(e^{5x} + x\right) \) helps in simplifying the expression so that we can apply L'Hopital's Rule. This transformation into a logarithmic form is a common technique when dealing with expressions raised to a power, as it reduces the complexity of calculus operations.

Exponential Function

The exponential function \(e^x\) is a vital mathematical concept with the base \(e\), a transcendental number approximately equal to 2.71828. It often appears in various natural processes such as growth and decay.

• One key property is that its derivative is itself: \(\frac{d}{dx}e^x = e^x\), which makes it particularly useful in calculus and differential equations.
• It relates naturally to its inverse, the natural logarithm \(\ln\), allowing seamless conversion between these forms.

This exercise utilizes the exponential function \(e^{5x}\), where raising \(e\) to increasing powers of \(x\) models growth. When evaluating limits involving exponential expressions, understanding its behavior as \(x\) approaches zero is crucial. The problem simplifies to finding the limit of an exponent, and ultimately, the power \(e^{-1}\) yields the final result of \(\frac{1}{e}\), leveraging these exponential properties.