limit

[metru]

Prove that:

\(\displaystyle {\lim }\limits_{x \to \infty } x^2 \cdot \left( {e^{\frac{1}
{x}} - e^{\frac{1}
{{x + 1}}} } \right) = 1\)

 

[tkhunny]

My first inclinaton is to substitute u = 1/x, since the new expression has a series expansion around u = 0.

 

[robbwrr]

\(\displaystyle {\lim }\limits_{x \to \infty } x^2 \left( {e^{\frac{1}
{x}} - e^{\frac{1}{{x + 1}}} } \right) = {\lim }\limits_{x \to \infty } x^2 \left( {e^{\frac{1}{x}-{\frac{1}{{x + 1}}} }-1 \right)e^{-\frac{1}{{x + 1}}}\)

\(\displaystyle = {\lim }\limits_{x \to \infty } x^2 \left( e^{\frac{1}{x(x+1)}}-1 \right)e^{-\frac{1}{{x + 1}}} ={\lim }\limits_{x \to \infty } x^2 \left( 1 +\frac{1}{x(x+1)} +O(\frac{1}{x^2(x+1)^2}) -1 \right)\)

\(\displaystyle = {\lim }\limits_{x \to \infty }\left( \frac{x^2}{x(x+1)} +x^2 \cdot O(\frac{1}{x^2(x+1)^2})\right )=1\)

 

[metru]

Prove that:

If \(\displaystyle \exists {\lim }\limits_{x \to \alpha } \omega \left( x \right) = 0\), then \(\displaystyle {\lim }\limits_{x \to \alpha } \frac{{e^{\omega \left( x \right)} - 1}}{{\omega \left( x \right)}} = 1\)

 

[stapel]

Please post new questions as new threads, not as replies to old threads, where they tend to be overlooked.

Thank you.

Eliz.

 

[galactus]

Yes, Stapel is correct. You ought to do that from here on out. New threads are always better.

Considering the condition \(\displaystyle \L\\\lim{\omega}(x)=0\), we have the

form \(\displaystyle \L\\\frac{0}{0}\)

Derivative of numerator and denominator.

\(\displaystyle \L\\\frac{{\omega}'(x)e^{{\omega}(x)}}{{\omega}'(x)}\)

=\(\displaystyle \L\\e^{{\omega}(x)}=e^{0}=1\)