# Question bb7dc

Apr 29, 2016

Assuming that ${\lim}_{x \rightarrow \infty} f \left(x\right) = 0$, use the fact that ${\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$ and the fact that the exponential function is continuous.

#### Explanation:

Note that ${\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$.

If ${\lim}_{x \rightarrow \infty} f \left(x\right) = 0$, then

${\lim}_{x \rightarrow \infty} \left(1 + f \left(x\right)\right) = {\lim}_{x \rightarrow \infty} {\left[{\left(1 + \frac{1}{\frac{1}{f} \left(x\right)}\right)}^{\frac{1}{f} \left(x\right)}\right]}^{f} \left(x\right)$

By the limit stated above, with $u = \frac{1}{f} \left(x\right)$, we have

${\lim}_{x \rightarrow \infty} u = \infty$, and so,

${\lim}_{x \rightarrow \infty} \left[{\left(1 + \frac{1}{\frac{1}{f} \left(x\right)}\right)}^{\frac{1}{f} \left(x\right)}\right] = {\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$

So

${\lim}_{x \rightarrow \infty} {\left(1 + f \left(x\right)\right)}^{g} \left(x\right) = {\lim}_{x \rightarrow \infty} {\left({\left[{\left(1 + \frac{1}{\frac{1}{f} \left(x\right)}\right)}^{\frac{1}{f} \left(x\right)}\right]}^{f} \left(x\right)\right)}^{g} \left(x\right)$

$= {\lim}_{x \rightarrow \infty} {\left({\left[e\right]}^{f} \left(x\right)\right)}^{g} \left(x\right)$

$= {\lim}_{x \rightarrow \infty} {e}^{f \left(x\right) g \left(x\right)}$

$= {e}^{{\lim}_{x \rightarrow \infty} f \left(x\right) g \left(x\right)}$

Notes

${\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$ is sometimes taken as the definition of $e$.

It can be verified using l'Hospital's Rule

$\ln \left({\left(1 + \frac{1}{u}\right)}^{u}\right) = u \ln \left(1 + \frac{1}{u}\right) = \ln \frac{1 + \frac{1}{u}}{\frac{1}{u}}$ which takes indeterminate form $\frac{- \infty}{\infty}$ as $x \rightarrow \infty$

L'Hospitals's rule asks us to eveluate

${\lim}_{u \rightarrow \infty} \frac{\frac{1}{1 + \frac{1}{u}} \left(- \frac{1}{u} ^ 2\right)}{- \frac{1}{u} ^ 2}$ which evaluates to $1$.

Since the limit of the $\ln$ is $1$, the limit of the expression is $e$.

Another Limit Result

If ${\lim}_{x \rightarrow a} f \left(x\right) = 1$ and ${\lim}_{x \rightarrow a} g \left(x\right) = \infty$,

then we can evaluate ${\lim}_{x \rightarrow a} {\left(f \left(x\right)\right)}^{g \left(x\right)}$ using ${\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$ to get

${\lim}_{x \rightarrow a} {\left(f \left(x\right)\right)}^{g \left(x\right)} = {e}^{{\lim}_{x \rightarrow a} \left(f \left(x\right) - 1\right) g \left(x\right)}$.

Outline of proof:

As $x \rightarrow a$,

we have $f \left(x\right) \rightarrow 1$, so

$\left(f \left(x\right) - 1\right) \rightarrow {0}^{+}$, and

$\frac{1}{f \left(x\right) - 1} \rightarrow \infty$

Therefore, as $x \rightarrow a$, (with $u = \frac{1}{f \left(x\right) - 1}$) we have

${\left(1 + \frac{1}{\frac{1}{f \left(x\right) - 1}}\right)}^{\frac{1}{f \left(x\right) - 1}} \rightarrow e$

We can conclude that

${\lim}_{x \rightarrow a} {\left(f \left(x\right)\right)}^{g \left(x\right)} = {\lim}_{x \rightarrow a} {e}^{\left(f \left(x\right) - 1\right) g \left(x\right)}$

$= {e}^{{\lim}_{x \rightarrow a} \left(f \left(x\right) - 1\right) g \left(x\right)}$

Although this equality holds for any $g \left(x\right)$, it is only interesting if ${\lim}_{x \rightarrow} g \left(x\right) = \infty \text{ or } - \infty$. Otherwise we get a triviality ${1}^{g} \left(x\right) = {e}^{0}$

Apr 30, 2016

Assume that as $x \to \infty$, both $f \left(x\right) \mathmr{and} g \left(x\right) \to 0$ and we are concerned about the limit of the ratio.. See explanation for proof/derivation, under this assumption

#### Explanation:

${\left(1 + f \left(x\right)\right)}^{\frac{1}{g} \left(x\right)} = {e}^{\frac{1}{g} \left(x\right)} \ln \left(1 + f \left(x\right)\right) = {e}^{\frac{1}{g} \left(x\right)} \left(f \left(x\right) + O \left({\left(f \left(x\right)\right)}^{2}\right)\right)$.

=e^(f(x)/g(x)(1+O(f(x))#

Now, take the limit as $x \to \infty$..