L'Hospital's Rule application with raised exponents. I am having a little trouble going about:
$$\lim_{x\to \infty} \left(\frac{14x}{14x+10}\right)^{10x}$$
Using $\ln$ properties we can bring down the $10x$ exponent and have:$$\ln y=10x\ln\left(\frac{14x}{14x+10}\right)$$
And from here I get stuck trying to apply L'Hospital's Rule to find the limit.
 A: Hint :
$$10x \ln \left(\frac{14x}{14x+10}\right)= \frac{\ln \left(\frac{14x}{14x+10}\right)}{\frac{1}{10x}}$$
A: HINT
We can use that
$$\left(\frac{14x}{14x+10}\right)^{10x}=\left(\frac{14x+10-10}{14x+10}\right)^{10x}=\left[\left(1-\frac{10}{14x+10}\right)^{\frac{14x+10}{10}}\right]^{\frac{10\cdot 10x}{14x+10}}$$
and refer to standard limits. 
A: Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.
$$\ln y=10x\ln\left(\frac{14x}{14x+10}\right)$$
$$\ln y=\frac{\ln\left(\frac{14x}{14x+10}\right)}{\frac{1}{10x}}$$
Now, you can continue.

As an alternative approach (both faster and easier), you can use $\lim_\limits{n \to \infty}\big(1+\frac{x}{n}\big)^n = e^x$. These expressions are easily manipulated to reach such a form.
Notice the numerator is $10$ less than the denominator, so $\frac{14x}{14x+14} = 1+\frac{-10}{14x+10}$. Hence, you get
$$\lim_{x\to \infty} \left(\frac{14x}{14x+10}\right)^{10x} = \lim_{x\to \infty} \Biggl[\left(1+\frac{-10}{14x+10}\right)^{14x+10}\Biggl]^{\frac{10x}{14x+10}}$$
A: Once you get to the logarithm, you need
$$
\lim_{x\to\infty}\frac{\ln\dfrac{14x}{14x+10}}{\dfrac{1}{10x}}
$$
which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $\ln14+\ln x-\ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get
$$
\lim_{x\to\infty}\frac{\dfrac{1}{x}-\dfrac{14}{14x+10}}{-\dfrac{1}{10x^2}}=
\lim_{x\to\infty}-10x^2\frac{14x+10-14x}{x(14x+10)}=\lim_{x\to\infty}-\frac{100x}{14x+10}
$$
When you have this limit, let me call it $l$, the one you started with is $e^l$.
A: Tips:
With equivalents, it's very fast:$$\ln\biggl(\frac{14x}{14x+10}\biggr)=\ln\biggl(1-\frac{10}{14x+10}\biggr)\sim_\infty -\frac{10}{14x+10}\sim_\infty -\frac{10}{14x}=-\frac 57,$$
so, as equivalence is compatible with multiplication/division,
$$10x\ln\biggl(\frac{14x}{14x+10}\biggr)\sim_\infty -\frac{50x}{7x}\to -\frac{50}7.$$
Warning: equivalence  is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.
