Find $\lim_{x\to\infty} x\exp(1/x)-x$. Is there a way to find the $$\lim_{x\to\infty} x\exp(1/x)-x$$ but without using L'Hospital's rule??
Let $1/x=t$. We are interested in 
$$\lim_{t\to 0^+}\frac{e^t-1}{t}.$$
We recognize this limit (with no restrictions on direction) as the derivative of $e^t$ at $t=0$.
$$xe^{1/x}-x=x\left(e^{1/x}-1\right)=\frac{e^{1/x}-1}{\frac1x}$$

As $\;x\to\infty\;$ both numerator and denominator tend to zero so by l'Hospital:

$$\lim_{x\to\infty}\frac{e^{1/x}-1}{\frac1x}\stackrel{\text{l'H}}=\lim_{x\to\infty}e^{1/x}=1$$