using L'Hospital solve $\lim_{x \to \infty} x - x^{2}\ln(1 + \frac{1}{x})$

Question

I can't get this to $ = \frac{0}{0}$ form so I can use l'Hospital rule $$\lim_{x \to \infty} x - x^{2}\ln\left(1 + \frac{1}{x}\right)$$

tips?

[EDIT] $$\lim_{x \to 0} \frac{1}{x} - \frac{\ln(1 + x)}{x^{2}}$$ second term $$\lim_{x \to 0} \frac{\ln(1 + x)}{x^{2}} = \lim_{x \to 0} \frac{\frac{1}{1 + x}}{2x} = 2\lim_{x \to 0} \frac{x}{x+1} = 0$$ first term $$ \lim_{x \to 0} \frac{1}{x} = \infty$$

is this okey?

Hint: for $x\to\infty$, $\ln(1+\frac1x)=\frac1x-\frac{1}{2x^2}+o(\frac{1}{x^2})$. — MickG, Mar 17 '16 at 20:41
Let $x=1/t$. We are interested in $\frac{t-\ln(1+t)}{t^2}$ as $t$ approaches $0$ from the right. — André Nicolas, Mar 17 '16 at 20:43

Mhenni Benghorbal · Accepted Answer · 2016-03-17T20:48:34.540

2

Letting $y=1/x$ gives

$$\lim_{y\to 0} \frac{y-\ln(1+y)} {y^2}. $$

I think you can see it now!

edited Mar 17 '16 at 20:48

answered Mar 17 '16 at 20:42

Mhenni Benghorbal

47,431

I edited my post, is it okay now? – tomtom Mar 17 '16 at 20:52
You should work with the form in my answer! – Mhenni Benghorbal Mar 17 '16 at 20:55
1

ok got it thanks – tomtom Mar 17 '16 at 21:00
You are welcome! – Mhenni Benghorbal Mar 17 '16 at 21:03
See also http://math.stackexchange.com/questions/1207250/limit-of-function-using-lh%C3%B4pitals-rule – Martin Sleziak Nov 09 '16 at 06:27

3SAT · Answer 2 · 2016-03-17T21:04:48.980

0

$t=1/x$ so $x=1/t$

$$\lim_{t\to 0} 1/t-(1/t)^2\ln(1+t) $$

$$=\lim_{t\to 0} \frac{t-\ln(1+t)}{t^2}$$

L'Hopital:

$$=\lim_{t\to 0} \frac{1-1/(t+1)} {2t} $$

$$=\frac 1 2\lim_{t\to 0} \frac{t} {\frac{(t+1)(t(t+1))}{t+1}}$$

assume $\frac{t}{t+1}\neq 0$

$$=\frac 1 2\lim_{t\to 0}\frac{1}{1+t}=\color{red}{\frac 1 2}$$

edited Mar 17 '16 at 21:04

answered Mar 17 '16 at 20:53

3SAT

7,512

using L'Hospital solve $\lim_{x \to \infty} x - x^{2}\ln(1 + \frac{1}{x})$

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