Need help solving a limit

[Pitagoras]

Hello!

I need to calculate the asymptote of \(\displaystyle f(x)=ln(e^{-x} -x)+3 \)

When trying to calculate \(\displaystyle m=\lim_{x \to -\infty}\frac{ f(x)}{x} \), I dont know how to do it.

Can anyone help me.

Thanks in advance.

Carlos Silva.
Portugal

 

[Ishuda]

Hello!

I need to calculate the asymptote of f(x)=ln(e^(-x)-x)+3

When trying to calculate m=lim f(x)/x when x tends to

, I dont know how to do it.

Can anyone help me.

Thanks in advance.

Carlos Silva.
Portugal

Are you allowed to use L'Hôpital's rule?

 

[Pitagoras]

No. I'm in secundary school in Portugal.

 

[Ishuda]

No. I'm in secundary school in Portugal.

What methods/formulas involving logs are you allowed to use?

 

[Pitagoras]

lim (e^x-1)/x=1 when x tends to 0

lim (ln(x+1))/x=1 when x tends to 0

lim (ln x) /x= 0 when x tends to +∞

lim e^x/x^p=+∞ when x tends to +∞


ln(axb) = ln(a) + ln(b)
ln(a/b) = ln(a)- ln(b)
ln x^p= p ln x

 

[Ishuda]

Hello!

I need to calculate the asymptote of f(x)=ln(e^(-x)-x)+3

When trying to calculate m=lim f(x)/x when x tends to

, I dont know how to do it.

Can anyone help me.

Thanks in advance.

Carlos Silva.
Portugal

First replace x with -x and g(x) with f(-x) and we then need
\(\displaystyle \lim_{x \to \infty}\frac{g(x)}{-x} = -\lim_{x \to \infty}\frac{g(x)}{x} = -\lim_{x \to \infty}\frac{ln(e^x + x) + 3}{x} = -\lim_{x \to \infty}\frac{ln(e^x + x)}{x} - \lim_{x \to \infty}\frac{3}{x}\)
\(\displaystyle =-\lim_{x \to \infty}\frac{ln(e^x + x)}{x} - 0 = -\lim_{x \to \infty}\frac{ln(e^x + x)}{x}\)

Can you go from there? If not show use what you have done/tell us where you are stuck so we can help further.

 

[Pitagoras]

First replace x with -x and g(x) with f(-x) and we then need
\(\displaystyle \lim_{x \to \infty}\frac{g(x)}{-x} = -\lim_{x \to \infty}\frac{g(x)}{x} = -\lim_{x \to \infty}\frac{ln(e^x + x) + 3}{x} = -\lim_{x \to \infty}\frac{ln(e^x + x)}{x} - \lim_{x \to \infty}\frac{3}{x}\)
\(\displaystyle =-\lim_{x \to \infty}\frac{ln(e^x + x)}{x} - 0 = -\lim_{x \to \infty}\frac{ln(e^x + x)}{x}\)

Can you go from there? If not show use what you have done/tell us where you are stuck so we can help further.

I already had tried that, but I got stuck there.
Then I tried to do another replace \(\displaystyle y=ln(e^x+x) \) and I got \(\displaystyle -\lim_{y \to \infty} \frac{y}{e^y-e^x} \)

Every replace that I try I always get a function with x and y.
I dont know what to do...

 

[Ishuda]

I already had tried that, but I got stuck there.
Then I tried to do another replace \(\displaystyle y=ln(e^x+x) \) and I got \(\displaystyle -\lim_{y \to \infty} \frac{y}{e^y-e^x} \)

Every replace that I try I always get a function with x and y.
I dont know what to do...

How about factoring (e^x + x) as e^x (1+x e^-x) and using
ln(ab) = ln(a) + ln(b)
and notice that
x e^-x = \(\displaystyle \dfrac{1}{\dfrac{e^x}{x^{-1}}}=\dfrac{1}{\dfrac{e^x}{x^{p}}}\)
where p=-1.

 

[Steven G]

I already had tried that, but I got stuck there.
Then I tried to do another replace \(\displaystyle y=ln(e^x+x) \) and I got \(\displaystyle -\lim_{y \to \infty} \frac{y}{e^y-e^x} \)

Every replace that I try I always get a function with x and y.
I dont know what to do...

Put the multiple 1/x as the power and then interchange the limit and ln. Go from there.

 

[Pitagoras]

How about factoring (e^x + x) as e^x (1+x e^-x) and using
ln(ab) = ln(a) + ln(b)
and notice that
x e^-x = \(\displaystyle \dfrac{1}{\dfrac{e^x}{x^{-1}}}=\dfrac{1}{\dfrac{e^x}{x^{p}}}\)
where p=-1.

Thank you very much. I got it! (I think p=1).