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I am trying to graph

$\ f(x) = ln(x)+\frac{1}{x}$

by hand.

Domain is $\ (0,\infty)$.

To find the x intercept I did

$\ -x \cdot ln(x) = 1 $

$\ x^x= \frac{1}{e} $

Which I realized has no solution. So the graph does not cross the x intercept. Because $\lim\limits_{x \to \infty }$ is clearly positive$\ \infty$, this means that the graph is never negative. I know the graph approaches ln(x) as x goes to infinity.

Goals:

  • Find global minimum
  • Find $\lim\limits_{x \to 0^+}$

Hints only please (no spoilers). Also I am taking calculus next year so if it cannot be done without calculus someone please let me know.

Edit: by no spoilers I mean no one just post a picture of the graph please.

Murey Tasroc
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2 Answers2

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You're making good progress. One additional way to approach this problem is by plotting a few points— that is, picking a few good values of $x$, computing $f(x)$, and plotting the result. This can help you get a feel for the shape of the curve.

You've also observed that for large values of $x$, the graph of $f(x)$ begins to resemble the graph of $\ln(x)$ because the $1/x$ contribution gets smaller and smaller. That's exactly right, and because $\ln(x)$ increases without limit, you know that the graph of $f(x)$ will increase (slowly) without limit as $x$ becomes large.

You can see something similar happen as $x$ approaches zero (while still being positive). I won't give away the answer, but you can try plugging in convenient shrinking values like $x=e^{-1}, e^{-2}, e^{-10}, e^{-100}$, and so on. (These are values that get closer and closer to zero. What happens to the value of $f(x)$?)

If you can figure that out, that will give you a sense of the overall shape of the curve. As for computing the minimum, I can't immediately think of any way to compute the global minimum without calculus, but I might be overlooking something. Good luck!

user326210
  • 17,287
  • Ahh I see. So $ (e^{-1} , e^{-2} , e^{-3} ...)$ would be (-1,-2,-3...) for lnx and (~2,~4,~8...) for $\frac{1}{x}$ so the $\ \frac{1}{x}$ is winning this fight. I guess that means it goes to infinity, and not a finite number (although I guess it would not make sense for it to be finite anyway because when adding neg/pos asymptotic graphs like these I realize one would always win out in the end unless they were exactly the same). Its different from something like $ x^x $, for example. +1 Thank you for the answer. I'm hoping for a minimum but maybe I'll have to wait for calculus to get that. – Murey Tasroc Jan 11 '18 at 04:19
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You could calculate the limit with pure algebra and knowledge of the limits of more basic functions.

\begin{align} \lim_{x \to 0^+} \ln(x)+\frac{1}{x} & = \lim_{x \to 0^+}\frac{x \ln(x)+1}x \\ & = \lim_{x \to 0^+} (x\ln(x)+1)\cdot\frac{1}{x} \\ & = \lim_{x \to 0^+} (x\ln(x)+1) \cdot \lim_{x \to 0^+}\frac{1}{x} \end{align}

Now we can evaluate each limit individually. I think this is pretty clear:

$$\lim_{x \to 0^+}\frac{1}{x}=\infty$$

And hopefully you've seen the graph of $y=\ln(x)$ before, so you know that

\begin{align} \lim_{x \to 0^+}x\ln(x)+1 \ge 0 \end{align}

since the $\log$ and $\ln$ functions are undefined for negative numbers and $0$. That means the limit function must be a positive number or $0$ itself.

Hence

\begin{align} \lim_{x \to 0^+} (x\ln(x)+1) \cdot \lim_{x \to 0^+}\frac{1}{x} & = k\cdot\infty \\ & = \infty \end{align}

for some positive integer $k$. This works well because even if $k=0$, the limit still holds.

As for the minimum, read up on First Principles Differentiation. It is just a formula which you can plug functions into, and equate to $0$ to find minimum and maximum values. Calculus isn't that hard, it is just very different. Good luck with your further studies!

Landuros
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