How to find $\lim _{x\to \infty \:}\left(\frac{x+3}{x+4}\right)^x$? I know the answer is 
$$
\lim _{x\to \infty \:}\ln\left(\frac{x+3}{x+4}\right)^x
=\lim _{x\to \infty \:}x\cdot\ln\left(\frac{x+3}{x+4}\right)
=\lim _{x\to \infty \:}\frac{\ln(x+3/x+4)}{1/x}=-1
$$
I know it has to do with L'Hopital rule, but I get the answer
$$
\lim _{x\to \infty \:}\frac{(x+4)/(x+3)}{-1/x^2},
$$
which is wrong.
Please help me.
 A: As an alternative
$$\left(\frac{x+3}{x+4}\right)^x=\left(\frac{x+4-1}{x+4}\right)^x=\left[\left(1-\frac{1}{x+4}\right)^{x+4}\right]^\frac{x}{x+4}$$
A: You may not have differentiated $\log_e((x+3)/(x+4))$ correctly.  As the logarithm of the quotient is the difference between the individual logarithms, thus 
$\log_e((x+3)/(x+4))=\log_e(x+3)-\log_e(x+4)$,
your numerator after applying l'H should be 
$(1/(x+3))-(1/(x+4))$.  
See where that gets you.
A: Using equivalents is much simpler than L'Hospital's rule: you have to find the limit of
$$x\ln\frac{x+3}{x+4}\; \text{ at }\infty. $$
Now, $\;\ln(1+u)\sim_0 u$, so 
$$x\ln\frac{x+3}{x+4}=x\ln\Bigl(1-\frac{1}{x+4}\Bigr)\sim_{x\to\infty}-\frac x{x+4} \sim_{x\to\infty}-\frac xx=-1.$$
A: $$ \left(\frac {x+3}{x+4}\right)^x = \left(1-\frac {1}{x+4}\right)^{(x+4)-4}\to e^{-1}$$
A: Set
$$
f(t)=\frac{1}{t}\ln\frac{3t+1}{4t+1}
$$
Since
$$
\left(\frac{x+3}{x+4}\right)^{\!x}=\exp(f(1/t))
$$
we can compute $\lim{t\to0}f(t)$ and if this exists and is $l$, the limit you're looking for is $e^l$.
Now
$$
\lim_{t\to0}\frac{\ln(3t+1)-\ln(4t+1)}{t}
$$
is the derivative at $0$ of the function $g(t)=\ln(3t+1)-\ln(4t+1)$; since
$$
g'(t)=\frac{3}{3t+1}-\frac{4}{4t+1}
$$
we have $g'(0)=3-4=-1$, so the limit you asked is $e^{-1}$.
Your method is very similar, but you got wrong the derivative of
$$
\ln\frac{x+3}{x+4}=\ln(x+3)-\ln(x+4)
$$
which is
$$
\frac{1}{x+3}-\frac{1}{x+4}=\frac{1}{(x+3)(x+4)}
$$
so you should have had
$$
\lim_{x\to\infty}\frac{\dfrac{1}{(x+3)(x+4)}}{-\dfrac{1}{x^2}}=
\lim_{x\to\infty}-\frac{x^2}{(x+3)(x+4)}=-1
$$
A: $\lim_{x \rightarrow \infty}\left (\dfrac{x^x(1+3/x)^x}{x^x(1+4/x)^x}\right )=$
$\dfrac {\lim_{x \rightarrow \infty}(1+3/x)^x}{\lim_{ x \rightarrow \infty}(1+4/x)^x}=\dfrac{e^3}{e^4}=e^{-1}.$
Note: 
Let $a >0$.
$(1+a/x)^x= ((1+\dfrac{1}{(\frac{x}{a})})^{\frac{x}{a}})^a.$
$y:=x/a.$
$\lim_{x \rightarrow \infty}(1+a/x)^x=$
$\lim_{y \rightarrow \infty}((1+1/y)^y)^a=e^a.$
A: $$ \lim \:_{x\to \:\infty \:}\left(\frac{x+3}{x+4}\right)^x\:$$
$$\lim \:_{x\to \:\infty \:}\:\:x\cdot ln\left(\frac{x+3}{x+4}\right)\:$$
$$ \lim \:_{x\to \:\infty \:}\:\:ln\left(\frac{x+3}{x+4}\right)\:\div \:\frac{1}{x}$$
 Now use L'Hopital's rule,
$$\lim \:_{x\to \:\infty \:}\:\:\frac{1}{\left(x+3\right)\left(x+4\right)}\:\div \:\frac{-1}{x^2}$$
$$ \lim \:_{x\to \:\infty \:}\:\:\frac{1}{\left(x+3\right)\left(x+4\right)} \cdot \:-x^2$$
$$\lim \:_{x\to \:\infty \:}\:\:\frac{-x^2}{x^2+7x+12}$$
Apply L'Hopital's rule again,
$$\lim _{x\to \infty \:}\left(\frac{-2x}{2x+7}\right)$$
and again,
$$\lim _{x\to \infty \:}\left(\frac{-2}{2}\right)=-1$$
