Limiting Logarithm: x→0+ log(x-1+√(x^2+1))-logx

[tylersmith7690]

Homework Statement

lim_x→0⁺ log( x - 1 +$\sqrt{x^2+1}$)-logx)

Note that log is the same as log_e same as ln. Couldn't find the three parallel lines symbol.

The Attempt at a Solution

I really don't know where to start with this one. Should I use the limit laws and bring the limit into the brackets to give

log( lim_x→0⁺ (x) - lim_x→0⁺ 1 + √(lim _x→0⁺x^2+1) -lim_x→0⁺ log x

I see that this so far is most likely wrong any tips would be appreciated.

 

[haruspex]

lim_x→0⁺ log( x - 1 +$\sqrt{x^2+1}$)-logx)

I assume you mean lim_x→0⁺ log( x - 1 +$\sqrt{x^2+1}$)-log(x) (i.e. the logs are not nested).
What seems natural when faced with log(a) - log(b)?
What can you do with $\sqrt{x^2+1}$ for small x?

 

[tylersmith7690]

Would i make the original equationin into

lim_x→0⁺ log ((x-1+$\sqrt{x^2+1}$)/x))

= lim_x→0⁺ log( 1 - $\frac{1}{x}$ + $\frac{\sqrt{x^2+1}}{x}$)

And then can i say that because log is continuous when x>0, that i can then put
log(lim_x→0⁺ 1 - lim_x→0⁺$\frac{1}{x}$ + lim _x→0⁺ $\frac{\sqrt{x^2+1}}{x}$)

Im not sure if that's right so far, I see the inside bracket becoming ∞.
Which means limit does not exist.

 

[HallsofIvy]

Would i make the original equationin into

lim_x→0⁺ log ((x+1+$\sqrt{x^2+1}$)/x))

= lim_x→0⁺ log( 1 + $\frac{1}{x}$ + $\frac{\sqrt{x^2+1}}{x}$)

I wouldn't. I would make it, rather
$$\log\left(\frac{x+1+\sqrt{x^2+ 1}}{x}\right)$$
Now, since "logarithm" is a continuous function, it becomes a question of
$$\lim_{x\to 0}\frac{x+1+\sqrt{x^2+ 1}}{x}$$
and since that "becomes "2/0" when x= 0, the limit does not exist. (More specifically, the limit is "infinity" which is just saying it does not exist in a specific way.)

And then can i say that because log is continuous when x>0, that i can then put
log(lim_x→0⁺ 1 + lim_x→0⁺$\frac{1}{x}$ + lim _x→0⁺ $\frac{\sqrt{x^2+1}}{x}$)

Im not sure if that's right so far, I see the inside bracket becoming ∞.
Which means limit does not exist.

 

[haruspex]

lim_x→0⁺ log ((x+1+$\sqrt{x^2+1}$)/x))

Shouldn't that be ... log ((x-1+...?

And then can i say that because log is continuous when x>0, that i can then put
log(lim_x→0⁺ 1 + lim_x→0⁺$\frac{1}{x}$ + lim _x→0⁺ $\frac{\sqrt{x^2+1}}{x}$)

You can, but doing that too soon will usually lead to indeterminate results, like +∞-∞. You need to get some cancellation done first. Can you write out an approximation for $\sqrt{x^2+1}$ that's valid for small x?

 

[jackmell]

Homework Statement

lim_x→0⁺ log( x - 1 +$\sqrt{x^2+1}$)-logx)

Note that log is the same as log_e same as ln. Couldn't find the three parallel lines symbol.

The Attempt at a Solution

I really don't know where to start with this one. Should I use the limit laws and bring the limit into the brackets to give

log( lim_x→0⁺ (x) - lim_x→0⁺ 1 + √(lim _x→0⁺x^2+1) -lim_x→0⁺ log x

I see that this so far is most likely wrong any tips would be appreciated.

. . . this has run-off into the ditch since early this morning and provides a nice example why it is so important to be utterly perfect in your math notation. Tell you what, no matter what you want to solve, how about we solve instead the perfectly-illustrated problem:

$$\lim_{x\to 0} \log\left(\frac{x-1+\sqrt{x^2+1}}{x}\right)$$

Put the limit inside. Justify it later. Looks like the indeterminate form $\frac{0}{0}$. How about L'Hospital's rule?

 

[tylersmith7690]

Sorry for the headache, my second attempt above at working out had a + instead of a - after the first x in the bracket. It should read (x-1 + ...)/x

Because the function log is continuous.

we get

lim_x→0⁺ ($\frac{x-1+\sqrt{}x^2+1}{x}$)

= $\frac{0-1+1}{0}$ which is $\frac{0}{0}$ indeterminate form.

Then use l'hopital rule

lim_x→0⁺ (1+ $\frac{1}{2}$(x²+1)^-1/2 .2x)/1

= lim_x→0⁺ $\frac{1+x}{\sqrt{}x^2+1}$

= $\frac{1}{1}$

 

[jackmell]

Sorry for the headache, my second attempt above at working out had a + instead of a - after the first x in the bracket. It should read (x-1 + ...)/x

Because the function log is continuous.

we get

lim_x→0⁺ ($\frac{x-1+\sqrt{}x^2+1}{x}$)

= $\frac{0-1+1}{0}$ which is $\frac{0}{0}$ indeterminate form.

Then use l'hopital rule

lim_x→0⁺ (1+ $\frac{1}{2}$(x²+1)^-1/2 .2x)/1

= lim_x→0⁺ $\frac{1+x}{\sqrt{}x^2+1}$

= $\frac{1}{1}$

That's not entirely right you know. You forgot the log:

$$\log\left(\lim_{x\to 0} \frac{1+\frac{x}{\sqrt{x^2+1}}}{1}\right)$$

And also, I am not 100% sure we can apply L'Hospital's rule for a right-handed limit. Is it continuous at zero? Maybe I'm in the ditch too. Also, the best advice I can give you this semester is to learn very good three things:

1. Latex
2. PF
3. Mathematica

I mean we need to get it right so I'll double-check my work in Mathematica. So I did, it's zero. Now, how can I prove that? And how can I write it so well so that there is no ambiguity? Latex. And how can I get excellent practice getting better? Hanging out in PF (follow rules carefully though).

 

[tylersmith7690]

So if I end up with

log(1) = 0,

Would that mean I was right?

or do i have to do some algebra first to solve the stuff inside the bracket?

 

[haruspex]

So if I end up with

log(1) = 0,

Would that mean I was right?

or do i have to do some algebra first to solve the stuff inside the bracket?

Yes, that's the right answer, but approximating √(1+x²) for small x using the usual expansion is a bit easier.

 

[tylersmith7690]

would i use the l'hopital rule in my working out? and what is the usual expansion?

 

[haruspex]

what is the usual expansion?

Binomial.

 

[vanhees71]

Usually the expansion of numerator and denominator in appropriate Taylor series is simpler than using de L'Hospital's rule, which is derived from this Taylor expansions anyway. As an example take what was discussed before (and please use LaTeX to type formulae; it's so much easier to read!):
$$\frac{x-1+\sqrt{x^2+1}}{x}=\frac{x-1+1+\mathcal{O}(x^2)}{x}=1+\mathcal{O}(x) \rightarrow 1 \quad \text{for} \quad x \rightarrow 0.$$