Value of the given limit.

Question

What is $$\lim _{ x\rightarrow 1 }{ \frac { x\log { \left( x \right) -x+1 } }{ \left( x-1 \right) \log { \left( x \right) } } } $$ Note I have used $\lim _{ x\rightarrow 0 }{ \frac { \log { \left( 1+x \right) } }{ x } } =1$ . So I wrote $\log(x)=\log(1+x-1)$ and hence I got $\frac{x^2-2x+1}{x-1}$ after cancelling I got the value as $0$. But the correct answer is $\frac{1}{2}$. Where's my mistake? Any hint. I don't want Lhospital rule or Taylor series if they aren't compulsory to work out the answer .

Answer 1

Now with editing this gives a solution using only the limit $\lim\limits_{y\to 0}\frac{e^{y}-1}{y}=1$. Start by setting $x=e^y$, you get $$\lim\limits_{y\to 0}\frac{ye^y-e^y+1}{(e^y-1)y}$$ Multiply by $\lim\limits_{y\to 0}\frac{e^y-1}{y}=1$ to get $$\lim\limits_{y\to 0}\frac{ye^y-e^y+1}{y^2}= \lim\limits_{y\to 0}\frac{e^y-1}{y}+\frac{1+y-e^y}{y^2}=1-L$$ Where $$L=\lim\limits_{y\to 0}\frac{e^y-1-y}{y^2}$$

Now using l'Hopital twice (while we try to think of another way) we see $L=\frac{1}{2}$.

To avoid l'Hopital

Write $$\lim\limits_{y\to 0}\frac{e^y-1-y}{y^2}=L$$

Then $$\lim\limits_{y\to 0}\frac{e^{-y}-1+y}{y^2}=L$$ adding $$\lim\limits_{y\to 0}\frac{e^{-y}+e^y-2}{y^2}=2L$$

Now on the other hand

$$\lim\limits_{y\to 0}\frac{e^{y}-1}{y}=1$$ and $$\lim\limits_{y\to 0}\frac{e^{-y}-1}{y}=-1$$

and multiplying them together $$\lim\limits_{y\to 0}\frac{2-e^{-y}-e^y}{y^2}=-1$$

It follows that $2L=1$.

Answer 2

HINT

Put $x-1=h $

As x goes to $1$, h goes to $0$. Then use expansion for $log(1+h)$ in numerator and result you stated in your attempt in denominator

$lim_{h \to0} \frac{(1+h)log(1+h) - h}{hlog(1+h)}$

$lim_{h \to0} \frac{(1+h)log(1+h) - h}{h^2}$

$lim_{h \to0} \frac{(1+h)(h-\frac{h^2}{2} + ...) - h}{h^2}$

Answer 3

It makes sense substituting $x=t+1$, so the limit becomes $$ \lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t\log(1+t)} $$ With a Taylor expansion it is quite easy: $$ \log(1+t)=t-\frac{t^2}{2}+o(t^2) $$ so $$ (1+t)\log(1+t)-t= (1+t)\left(t-\frac{t^2}{2}+o(t^2)\right)-t= \frac{t^2}{2}+o(t^2) $$ and the denominator can be written $t\log(1+t)=t(t+o(t))=t^2+o(t^2)$.

Therefore $$ \lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t\log(1+t)} = \lim_{t\to0}\frac{\dfrac{t^2}{2}+o(t^2)}{t^2+o(t^2)}=\frac{1}{2} $$

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You can also use l'Hôpital, but first rewrite the limit as $$ \lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t\log(1+t)} = \lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t^2}\,\frac{t}{\log(1+t)} $$ The limit of the first factor becomes $$ \lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t^2}\;\overset{\mathrm{(H)}}{=}\; \lim_{t\to0}\frac{\log(1+t)}{2t}=\frac{1}{2} $$ The second factor has limit $1$.

Answer 4

The problem is quite easy using Taylor and L’Hôpital but the O.P. wishes this way forbidden so the question becomes, at least for me, something difficult. Anyway, here a method of approximation to solve the question (especially when it is knowing that the limit is $\frac 12$):

Let $E$ be the given expression. We have $$E=\frac{x}{x-1}-\frac{1}{\ln(x)}$$ Put $x=t+1$ so we have $$E=\frac{t+1}{t}-\frac{1}{\ln(t+1)}=1+f(t)$$ where $$f(t)=\frac{1}{t}-\frac{1}{\ln(1+t)}$$ The domain of $f$ is $(-1,0)\cup (0,\infty)$ however if the searched limit exists then prolonging by continuity we know the function $f$ can be considered as continuous on $(-1, \infty)$. The derivative is $$f’(t)=-\frac{1}{t^2}+\frac{1}{\ln^2(1+t)}\gt 0$$ so the function is deduced to be increasing on its domain. Now, calculating some values in the neighborhood of $t=0$ $$ f(0.01)\approx -0.499170\\ f(0.001)\approx -0.499916\\f(0.0001)\approx -0.499991\\ f(-0.01)\approx -0.500837\\f(-0.001)\approx -0.500083\\f(-0.0001)\approx -0.500008$$ Knowing that the limit exists, that $f$ is continuous for $x\gt -1$ and increasing, we can conclude, that this limit indeed is equal to ${\frac {-1}{2}}$. Thus $$\lim_{t\to 0}E=1-\frac 12=\color{red}{\frac 12}$$