find $y= \lim_\limits{t\to\infty} \left({{t-3}\over {2t+1}}\right)^t$ 
find $y= \lim_\limits{t\to\infty} \left({{t-3}\over {2t+1}}\right)^t$

My attempt:
$$\ln y= \ln{\left({{t-3}\over {2t+1}}\right)^t}$$
$$\lim_\limits{t\to\infty} \ln{y} = \lim_\limits{t\to\infty} t(\ln{(t-3)}- \ln{(2t+1)})$$
True? and how can I complete?
Thanks
 A: Compute this way:
$$\lim_{t\to\infty}t\cdot\ln \left(\frac{t-3}{2t+1}\right)=\lim_{t\to\infty}\frac{\ln\left(\frac{t-3}{2t+1}\right)}{\frac1t}.$$
Here the numerator tends to $\ln\frac12<0$ and the denominator tends to zero through positive numbers. So the limit is $-\infty$.
A: The limit is zero. To see this, note that
$$
\frac{t-3}{2t+1} = \frac{1-3/t}{2+1/t}
$$
converges to $1/2,$ and thus is less than $2/3$ in absolute value for $t\ge t_0,$ for some $t_0>0.$ But
$$
\left(\frac{2}{3}\right)^t
$$
converges to 0.
A: It is : 
$$\lim_{t\to \infty}\bigg(\frac{t-3}{2t+1}\bigg)=\frac{1}{2}$$
and :
$$\lim_{t \to \infty} \bigg(\frac{1}{2}\bigg)^t=0$$
hence : 
$$y= \lim_{t\rightarrow\infty} \left({{t-3}\over {2t+1}}\right)^t=0$$
A: Note that $\ln(y)$ can be rewritten as
$$
\ln(y) = \lim_{t\to\infty}\frac{\ln\Big(\frac{t-3}{2t+1}\Big)}{1/t}.
$$
However, by continuity of ln, the numerator tends to $\ln(1/2) = -\ln(2)$ as $t\to\infty$ and the denominator tends to $0$ as $t\to\infty$. Hence $\ln(y) = -\infty$. Now deduce that $y = 0$.
A: Try to use that $\frac{t-3}{2t+1}\rightarrow\frac{1}{2}.$
A: $$
y= \lim_{t\rightarrow\infty} \left({{t-3}\over {2t+1}}\right)^t \\
= \lim_{t\rightarrow\infty} \left(\frac12 \cdot {{t-3}\over {t+0.5}}\right)^t \\
= \lim_{t\rightarrow\infty} \left(\frac12 \cdot (1 - {{3.5}\over {t+0.5}})\right)^t 
$$
Let $(t+0.5)/3.5 = x$ then we have 
$$
y= \lim_{x\rightarrow\infty} \left(\frac12 \cdot (1 - \frac1x)\right)^{3.5 x - 0.5}
$$
Now use that for large x, $(1-1/x)^x \to e^{-1}$, so we have 
$$
y= \lim_{x\rightarrow\infty} \left(\frac12 \cdot e^{-1/x}\right)^{3.5 x - 0.5} \\
\lim_{x\rightarrow\infty}  \left(\frac12 \right)^{3.5 x - 0.5} \cdot \left(e^{-3.5 +0.5/x}\right)  \\
= e^{-3.5} \lim_{x\rightarrow\infty}  \left(\frac12 \right)^{3.5 x - 0.5} = 0
$$
A: Set $h=1/t$
$$y= \lim_{t\rightarrow\infty} \left({{t-3}\over {2t+1}}\right)^t =\lim_{h\rightarrow0^+ } \left({{1-3h}\over {2+h}}\right)^{1/h} \\= \lim_{h\rightarrow0 } \left({{1-3h}\over {2+h}}\right)^{1/h} \\=\lim_{h\rightarrow0^+ }\exp\left(3\frac{\ln(1-3h)}{3h} -\frac{\ln(h +2)}{h}\right)  = 0$$
Since $$ \lim_{h\rightarrow0 } -\frac{\ln(h +2)}{h} =-\infty$$
and $$ \lim_{h\rightarrow0 } \frac{\ln(1-3h)}{-3h} = 1$$
