Find $\lim _{x\to \infty }\left(x\left(\ln\left(x+1\right)-\ln x\right)\right) $

Question

Find

$$\lim _{x\to \infty }x\Big(\ln\left(x+1\right)-\ln x\Big)$$

Here's how I do it:

$$x\Big(\ln(x+1)-\ln x\Big) = x\Bigg(\ln(x(1+\frac{1}{x}) - \ln x\Bigg)$$ $$x\Big(\ln x + \ln(1+\frac{1}{x}\Big)-\ln x =x\ln\Big(1+\frac{1}{x}\Big)$$ $$\ln\Bigg( \Big(1+\frac{1}{x}\Big)^x \Bigg) \rightarrow \ln 1^\infty = 0$$

What am I doing wrong? The answer is supposed to be $1$, but I get $0$.

score 4 · Answer 1 · answered Feb 01 '15 at 18:12

4

$1^\infty\neq 1$. It is a case you have to investigate. Here, $$ \lim_{x\to +\infty}\bigl(1+1/x\bigr)^x = e. $$ This should be derived in your book.

answered Feb 01 '15 at 18:12

mickep

19,962

score 1 · Accepted Answer · answered Feb 01 '15 at 18:13

1

Just write it as

$$ \lim_{x \to \infty} x\ln(1+1/x) = \lim_{t \to 0} \frac{ \ln(1+t)}{t}=\dots. $$

I think you can finish it!

answered Feb 01 '15 at 18:13

Mhenni Benghorbal

47,431

How come they are equal? – lawls Feb 01 '15 at 18:23
@lawls: I used $\ln(a)-\ln(b)= \ln(a/b) $ then I put $ t=1/x $. – Mhenni Benghorbal Feb 01 '15 at 18:24
Oh I see, but how come you changed to x->0 instead of infinity? – lawls Feb 01 '15 at 18:28
Since we put $t=1/x$ then as $x\to \infty$ we will have $t\to 0$. – Mhenni Benghorbal Feb 01 '15 at 18:29

score 1 · Answer 3 · answered Feb 01 '15 at 18:13

1

Hint: Let $y = \frac{1}{x}$ then $$\lim_{y\to 0} \frac{\ln( 1 + y)}{y} = \ldots$$

Can you get it from here?

Note: $\lim_{x\to \infty} (1 + \frac{1}{x})^x = e$ and $\ln e = 1$.

answered Feb 01 '15 at 18:13

Aaron Maroja

17,571

score -1 · Answer 4 · answered Feb 01 '15 at 18:15

-1

look at $$x\left(\ln(x+1) - \ln(x) \right) = x \ln\left(\dfrac{x+1}{x}\right) = x\left( \ln(1 + 1/x) \right) = x\left(1/x -1/(2x^2) + \cdots \right) = 1$$ for $x \to \infty$

answered Feb 01 '15 at 18:15

abel

29,170

Find $\lim _{x\to \infty }\left(x\left(\ln\left(x+1\right)-\ln x\right)\right) $

4 Answers4

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