Asymptotes

Bolzano

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Determine asymptotes of the following function:
[math]f(x) = \frac{x - 2}{e^{1/x}}[/math]I've managed to find the vertical asymptote [imath]x = 0[/imath], but I'm struggling with oblique one. It's equation is [imath]y = ax + b[/imath] and I calculated that [imath]a = 1[/imath] using formula
[math]a = \lim_{x \to \infty} \frac{f(x)}{x}[/math]Now, I have a problem finding b. I used formula
[math]b = \lim_{x \to \infty} (f(x) - ax)[/math]and got that
[math]b = \lim_{x \to \infty} \bigg(\frac{x - 2}{e^{1/x} } - x\bigg)[/math]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.
 
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blamocur

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

Bolzano

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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 [imath]\frac{0}{0}[/imath].
 

blamocur

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

Bolzano

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After scratching my head for a while I could not come up with a proof that [imath]\lim\limits_{u\rightarrow 0} \frac{e^u-1}{u} = -1[/imath] 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 [imath]\lim_{x \to 0} \frac{e^x - 1}{x} = 1[/imath].

Let [imath]e^x - 1 = u \Leftrightarrow x = \ln(u + 1)[/imath]. Then
[math]\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[/math]
 
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Dr.Peterson

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

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

Let [imath]e^x - 1 = u \Leftrightarrow x = \ln(u + 1)[/imath]. Then
[math]\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[/math]
Another approach is just to recognize [imath]\lim_{x \to 0} \frac{e^x - 1}{x}[/imath] as the derivative of [imath]e^x[/imath] at [imath]x=0[/imath], assuming you've seen that.
 

blamocur

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

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

Let [imath]e^x - 1 = u \Leftrightarrow x = \ln(u + 1)[/imath]. Then
[math]\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[/math]
Really like your solution!
 
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