Graphing natural log of x plus one over x

Question

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.

Answer 1

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!

Answer 2

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!