calculate the following limit $ \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] $ Could you please help me to calculate the following limit
$$\mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right]$$
without Hôpital (otherwise)?
 A: Put $t=1+u$ so that
$$\lim_{t \to 1} \frac{(1+t)\ln t}{t\ln t - t + 1}=\lim_{u\to 0}\underbrace{\frac{(2+u)\ln(1+u)}{(1+u)\ln(1+u)-u} }_{f(u)}$$
Using the Taylor's expansion $\ln(1+u)\sim u$ for $u\to 0$ we have
$$
f(u)\sim \frac{(2+u)u}{(1+u)u-u} =1+\frac{2}{u} \quad\text{for }u \to 0
$$
so that $f(u)\to +\infty$ as $u\to 0^+$ and $f(u)\to -\infty$ as $u\to 0^-$.
Thus
$$
\lim_{t \to 1^{\pm}} \frac{(1+t)\ln t}{t\ln t - t + 1}=\pm\infty
$$
A: Set $t=1+u \enspace(u\to 0)$. Then $\ln t \sim u $. Express the fraction with $u$ and take equivalents for numerator and denominator.
A: Hint
If you already know Taylor series, you could use them building at $t=1$. So, $$\log(t)=(t-1)-\frac{1}{2} (t-1)^2+O\left((t-1)^3\right)$$ So, the numerator is $$2 (t-1)+\frac{1}{6} (t-1)^3+O\left((t-1)^4\right)$$ and denominator is $$\frac{1}{2} (t-1)^2-\frac{1}{6} (t-1)^3+O\left((t-1)^4\right)$$
Now, consider the ratio of both.
I am sure that you can take from here.
A: I provide computation without hospital's rule nor Taylor series. First, re-write
\begin{eqnarray*}
\frac{(2+u)\ln (1+u)}{(1+u)\ln (1+u)-u} &=&\frac{(2+u)\ln (1+u)}{(1+u)\left(
\ln (1+u)-u\right) +u^{2}} \\
&& \\
&=&\frac{\left( 1+\dfrac{2}{u}\right) \left( \dfrac{\ln (1+u)}{u}\right) }{%
(1+u)\left( \dfrac{\ln (1+u)-u}{u^{2}}\right) +1}
\end{eqnarray*}
Now we use the following standard limits
\begin{equation*}
\lim_{u\rightarrow 0}\dfrac{\ln (1+u)}{u}=1,\ \ \ \ \ and\ \ \ \ \ \
\lim_{u\rightarrow 0}\dfrac{\ln (1+u)-u}{u^{2}}=-\frac{1}{2}
\end{equation*}
one gets
\begin{equation*}
\lim_{u\rightarrow 0^{+}}\frac{(2+u)\ln (1+u)}{(1+u)\ln (1+u)-u}%
=\lim_{u\rightarrow 0^{+}}\frac{\left( 1+\dfrac{2}{u}\right) \left( \dfrac{%
\ln (1+u)}{u}\right) }{(1+u)\left( \dfrac{\ln (1+u)-u}{u^{2}}\right) +1}=%
\frac{\left( 1+\infty \right) \left( 1\right) }{(1+0)\left( -\frac{1}{2}%
\right) +1}=+\infty 
\end{equation*}
\begin{equation*}
\lim_{u\rightarrow 0^{-}}\frac{(2+u)\ln (1+u)}{(1+u)\ln (1+u)-u}%
=\lim_{u\rightarrow 0^{-}}\frac{\left( 1+\dfrac{2}{u}\right) \left( \dfrac{%
\ln (1+u)}{u}\right) }{(1+u)\left( \dfrac{\ln (1+u)-u}{u^{2}}\right) +1}=%
\frac{\left( 1-\infty \right) \left( 1\right) }{(1+0)\left( -\frac{1}{2}%
\right) +1}=-\infty .
\end{equation*}
