Find $\lim_{x\to\infty} x\exp(1/x)-x$.

Question

Is there a way to find the $$\lim_{x\to\infty} x\exp(1/x)-x$$ but without using L'Hospital's rule??

Answer 1

Since $x\to \infty$ we can use Taylor Series for $e^{1/x}$:

$$e^{1/x}\approx 1 + \frac{1}{x}$$

Thence

$$x\left(1 + \frac{1}{x}\right) - x$$

So the limit is

$$\boxed{1}$$

Answer 2

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 $t=0$.

Answer 3

$$xe^{1/x}-x=x\left(e^{1/x}-1\right)=\frac{e^{1/x}-1}{\frac1x}$$

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

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

Answer 4

As already said in answers, Taylor series is the most efficient way to solve the problem.

What is interesting is that, using one more term, you could see both the limit and how it is approached. $$e^{\frac{1}{x}}=1+\frac{1}{x}+\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ $$xe^{\frac{1}{x}}-x=1+\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ So, the limit is $1$ and it is approached from above.