MHB Derivative and Limit of an Exponential Function

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SUMMARY

The discussion focuses on finding the derivative and limit of the function \(f(x)=\left ( e^{x}+x \right )^{\frac{1}{x}}\). Participants agree on using the natural logarithm to simplify the differentiation process, leading to the expression \(\ln(y)=\frac{1}{x}\ln\left(e^x+x\right)\). For the limit as \(x\) approaches infinity, they apply L'Hôpital's Rule to resolve the indeterminate form \(\frac{\infty}{\infty}\), ultimately confirming that the limit evaluates to \(e\).

PREREQUISITES
  • Understanding of exponential functions and their properties
  • Familiarity with natural logarithms and their applications in calculus
  • Knowledge of implicit differentiation and the chain rule
  • Proficiency in applying L'Hôpital's Rule for limit evaluation
NEXT STEPS
  • Study the application of L'Hôpital's Rule in various indeterminate forms
  • Learn more about implicit differentiation techniques in calculus
  • Explore the properties of exponential functions and their limits
  • Review the chain rule and its implications in differentiation
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in advanced differentiation techniques and limit evaluation of exponential functions.

Yankel
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Hello all,

I have a complicated function:

\[f(x)=\left ( e^{x}+x \right )^{^{\frac{1}{x}}}\]

I need to find it's derivative and it's limit when x goes to infinity.

As for the derivative, I thought maybe to use LN, so that I can get rid of the exponent, am I correct?

How should I approach the limit ?

Thank you !
 
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Yes, I would agree that to find the derivative, we can write:

$$y=\left(e^x+x\right)^{\frac{1}{x}}$$

Then take the natural log of both sides:

$$\ln(y)=\frac{1}{x}\ln\left(e^x+x\right)$$

Now implicitly differentiate.

For the limit, logs can come to our aid as well...begin by writing:

$$L=\lim_{x\to\infty}\left(\left(e^x+x\right)^{\frac{1}{x}}\right)$$

Upon taking the natural log of both sides, we obtain:

$$\ln(L)=\lim_{x\to\infty}\left(\frac{\ln\left(e^x+x\right)}{x}\right)$$

This is the indeterminate form $$\frac{\infty}{\infty}$$ and so we may apply L'Hôpital's Rule.

Can you proceed?
 
Is the final answer of the limit e ? I have proceeded and got that the right hand side is 1.

What do you mean to derive implicitly?
 
Yankel said:
Is the final answer of the limit e ? I have proceeded and got that the right hand side is 1.

Yes, I got the same result. :D

Yankel said:
What do you mean to derive implicitly?

Implicit differentiation is nothing more than a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions $y$ written EXPLICITLY as functions of $x$. However, some functions $y$ are written IMPLICITLY as functions of $x$, as is the case once we took the natural logs of both sides. So, we differentiate, bearing in mind that the chain rule must be used with $y$ since it is a function of $x$. :D
 

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