# Which values of x this function is continuous? (x non-negative real number and n->infinity)

## $f \left(x\right) = \lim \log \frac{{e}^{n} + {x}^{n}}{n}$

Jan 5, 2018

For $x \in \left(0 , e\right)$, write the function under the limit as:

$\log \frac{{e}^{n} + {x}^{n}}{n} = \log \frac{{e}^{n} \left(1 + {\left(\frac{x}{e}\right)}^{n}\right)}{n}$

and using the properties of logarithms:

$\log \frac{{e}^{n} + {x}^{n}}{n} = \frac{\log \left({e}^{n}\right) + \log \left(1 + {\left(\frac{x}{e}\right)}^{n}\right)}{n}$

$\log \frac{{e}^{n} + {x}^{n}}{n} = 1 + \frac{\log \left(1 + {\left(\frac{x}{e}\right)}^{n}\right)}{n}$

Now, as $\frac{x}{e} < 1$ we have ${\lim}_{n \to \infty} {\left(\frac{x}{e}\right)}^{n} = 0$, so $f \left(x\right) = 1$

For $x = e$, we have that:

$\log \frac{{e}^{n} + {x}^{n}}{n} = \log \frac{2 {e}^{n}}{n} = \frac{\log 2 + n}{n}$

and again $f \left(x\right) = 1$

For $x > e$ write the function as:

$\log \frac{{e}^{n} + {x}^{n}}{n} = \log \frac{{x}^{n} \left({\left(\frac{e}{x}\right)}^{n} + 1\right)}{n}$

$\log \frac{{e}^{n} + {x}^{n}}{n} = \frac{n \log x + \log \left({\left(\frac{e}{x}\right)}^{n} + 1\right)}{n}$

$\log \frac{{e}^{n} + {x}^{n}}{n} = \log x + \log \frac{{\left(\frac{e}{x}\right)}^{n} + 1}{n}$

We have now $\frac{e}{x} < 1$ and ${\lim}_{n \to \infty} {\left(\frac{e}{x}\right)}^{n} = 0$, so $f \left(x\right) = \log x$

We can conclude that the function $f \left(x\right)$ is:

$f \left(x\right) = \left\{\begin{matrix}1 \text{ for " x <=e \\ logx " for } x > e\end{matrix}\right.$

and is therefore continuous for every $x \in \mathbb{R}$

Jan 5, 2018

It is continuous for all non-negative real $x$

#### Explanation:

It looks like the function is:

$f \left(x\right) = {\lim}_{n \rightarrow \infty} \ln \frac{{e}^{n} + {x}^{n}}{n}$

In the following $x$ always represents a non-negative real number.

For every $x$ ${\lim}_{n \rightarrow \infty} \left({e}^{n} + {x}^{n}\right) = \infty$ so the defining limit above has indeterminate form $\frac{\infty}{\infty}$.

Apply l"Hospital's Rule (differentiate with respect to $n$) and find

${\lim}_{n \rightarrow \infty} \frac{{e}^{n} + {x}^{n} \ln x}{{e}^{n} + {x}^{n}}$

Case 1 For $x < e$ so $\frac{x}{e} < 1$ divide numerator and denominator by $e$ to get

(1+(x/e)^nlnx)/(1+(x/e)^n whose limit at $\infty$ is $1$.

Case 2 For $x = e$, we have ${\lim}_{n \rightarrow \infty} \frac{{e}^{n} + {e}^{n}}{{e}^{n} + {e}^{n}} = {\lim}_{n \rightarrow \infty} 1 = 1$

Case 3 For $x > e$ so $\frac{e}{x} < 1$ divide numerator and denominator by $e$ to get

$\frac{{\left(\frac{e}{x}\right)}^{n} + \ln x}{{\left(\frac{e}{x}\right)}^{n} + 1}$ whose limit at $\infty$ is $\ln x$.

So,

$f \left(x\right) = \left\{\begin{matrix}1 & \text{if" & x<=e \\ lnx & "if} & x > e\end{matrix}\right.$ which is continuous at all $x \ge 0$