# 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.

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:

$$\ln y = \pi - e$$

$$y = e^{\pi - e}$$

• thanks bro, but how do you know that the numerator tends to 0, shouldnt it tend to infinity? – hazard Nov 17 '15 at 11:05
• @hazard i'll edit my answer – Varun Iyer Nov 17 '15 at 11:05
• @hazard did that help you understand it a bit better? – Varun Iyer Nov 17 '15 at 11:08
• ok thanks, my concept was wrong. thanks for the detailed answer. – hazard Nov 17 '15 at 11:09

Hint: The expression equals

$$\left [\left(1+ \frac{\pi -e}{x+e}\right)^{x+e}\right]^{x/(x+e)}.$$