Asymptotes

[Bolzano]

Determine asymptotes of the following function:
$$f(x) = \frac{x - 2}{e^{1/x}}$$I've managed to find the vertical asymptote $x = 0$, but I'm struggling with oblique one. It's equation is $y = ax + b$ and I calculated that $a = 1$ using formula
$$a = \lim_{x \to \infty} \frac{f(x)}{x}$$Now, I have a problem finding b. I used formula
$$b = \lim_{x \to \infty} (f(x) - ax)$$and got that
$$b = \lim_{x \to \infty} \bigg(\frac{x - 2}{e^{1/x} } - x\bigg)$$which I don't know how to solve.

P.S.
I don't speak English very well and it's my first time asking question here, so sorry if there are some mistakes.

 

[blamocur]

I would replace 1/x with a new variable and use L'Hopital's rule.

 

[Bolzano]

I would replace 1/x with a new variable and use L'Hopital's rule.

I'm not allowed to use L'Hopital's rule.

 

[blamocur]

Even without L'Hopital's rule switching to a variable which tends to 0 seems to make it more manageable.

 

[Bolzano]

Even without L'Hopital's rule switching to a variable which tends to 0 seems to make it more manageable.

I've tried that and still wasn't able do it. I always end up with $\frac{0}{0}$.

 

[blamocur]

I've tried that and still wasn't able do it. I always end up with $\frac{0}{0}$.

Which methods for solving 0/0 type limits are you familiar with ?

 

[blamocur]

After scratching my head for a while I could not come up with a proof that $\lim\limits_{u\rightarrow 0} \frac{e^u-1}{u} = -1$ without using either L'Hopital's rule or Tailor's series.

 

[Bolzano]

After scratching my head for a while I could not come up with a proof that $\lim\limits_{u\rightarrow 0} \frac{e^u-1}{u} = -1$ without using either L'Hopital's rule or Tailor's series.

I solved it! Thank you for giving me an idea.

Here's how I proved that $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$.

Let $e^x - 1 = u \Leftrightarrow x = \ln(u + 1)$. Then
$$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{u \to 0} \frac{u}{\ln(u + 1)} = \lim_{u \to 0} \frac{1}{\frac{1}{u}\ln(u + 1)} = \lim_{u \to 0} \frac{1}{\ln(u + 1)^{1/u}} = \frac{1}{\ln e} = 1$$

 

[Dr.Peterson]

I solved it! Thank you for giving me an idea.

Here's how I proved that $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$.

Let $e^x - 1 = u \Leftrightarrow x = \ln(u + 1)$. Then
$$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{u \to 0} \frac{u}{\ln(u + 1)} = \lim_{u \to 0} \frac{1}{\frac{1}{u}\ln(u + 1)} = \lim_{u \to 0} \frac{1}{\ln(u + 1)^{1/u}} = \frac{1}{\ln e} = 1$$

Another approach is just to recognize $\lim_{x \to 0} \frac{e^x - 1}{x}$ as the derivative of $e^x$ at $x=0$, assuming you've seen that.

 

[blamocur]

I solved it! Thank you for giving me an idea.

Here's how I proved that $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$.

Let $e^x - 1 = u \Leftrightarrow x = \ln(u + 1)$. Then
$$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{u \to 0} \frac{u}{\ln(u + 1)} = \lim_{u \to 0} \frac{1}{\frac{1}{u}\ln(u + 1)} = \lim_{u \to 0} \frac{1}{\ln(u + 1)^{1/u}} = \frac{1}{\ln e} = 1$$

Really like your solution!