Derivative and Limit of an Exponential Function

[Yankel]

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 !

 

[MarkFL]

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

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

Then take the natural log of both sides:

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

Now implicitly differentiate.

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

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

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

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

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

Can you proceed?

 

[Yankel]

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?

 

[MarkFL]

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

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