What is the value of $\lim_{x\to \infty}({x+\pi\over x+e})^x$ $$\lim_{x\to \infty}\left({x+\pi\over x+e}\right)^x$$
Can someone help me with this question. I tried exponentiating the function but I am not able to get in the indeterminate form to apply L'Hopitals rule.
 A: We have the limit:
$$y = \lim_{x\to \infty}\left({x+\pi\over x+e}\right)^x$$
Taking the natural log of both sides:
$$\ln y = \lim_{x\to \infty}x\ln\left({x+\pi\over x+e}\right) = \lim_{x\to \infty}\frac{\ln\left({x+\pi\over x+e}\right)}{\frac{1}{x}}$$
Now see that if we compute the limit, we get an indeterminate form, $\frac{0}{0}$ (We divide the natural log by $x$ on the numerator and the denominator):
$$\lim_{x\to \infty}\frac{\ln\left({x+\pi\over x+e}\right)}{\frac{1}{x}} = \lim_{x\to \infty}\frac{\ln\left({\frac{1+\frac{\pi}{x}}{1+\frac{e}{x}}}\right)}{\frac{1}{x}} = \frac{\ln 1}{0} = \frac{0}{0}$$
Thus, we can apply L'Hopital's rule:
$$\lim_{x\to \infty}\frac{\ln\left({x+\pi\over x+e}\right)}{\frac{1}{x}} = \frac{\frac{e - \pi}{(x+e)(x+\pi)}}{-\frac{1}{x^2}} = -\frac{(e-\pi)x^2}{(x+e)(x+\pi)} = \frac{(\pi - e)x^2}{(x+e)(x+\pi)}$$
We can compute the limit:
$$\lim_{x\to \infty}\frac{(\pi - e)x^2}{(x+e)(x+\pi)} = \lim_{x\to \infty}\frac{\pi - e}{1 + \frac{e + \pi}{x} + \frac{e\pi}{x^2}} = \pi - e$$
Thus, we get that:
$$\ln y = \pi - e$$
$$y = e^{\pi - e}$$
A: Hint: The expression equals
$$\left [\left(1+ \frac{\pi -e}{x+e}\right)^{x+e}\right]^{x/(x+e)}.$$
