Continuity of Functions with Limits to Infinity

[Felafel]

hi everyone, I've found this exercise on a textbook and it doesn't resemble any exercise I've seen before. I just want to know how to proceed, you don't have to solve it for me :)

Homework Statement

Study the continuity of the following functions, defined by:

1- f(x) = lim (n^x-n^-x)/(n^x+n^-x) x∈|R
n->+∞


2- f(x) = lim [ln(e^n+x^n)]/n x∈|R
n->+∞

The Attempt at a Solution

A function is continuos if its limit L exists and it equals f(L).
But the limit here is to +∞!
So, after computing the two limits for the given n->+∞, how do I go on studying the finction?

Many thanksss

 

[tiny-tim]

Hi Felafel!

So, after computing the two limits for the given n->+∞, how do I go on studying the finction?

You'll get the value of f(x) for various values of x.

Draw the graph (in your head, if it's easy), and it should be obvious whether it's continuous!

 

[Felafel]

Hi Felafel!


You'll get the value of f(x) for various values of x.

Draw the graph (in your head, if it's easy), and it should be obvious whether it's continuous!

thank you :)!
just.. random values?

 

[tiny-tim]

just.. random values?

yup!

usually works! ​

 

[HallsofIvy]

hi everyone, I've found this exercise on a textbook and it doesn't resemble any exercise I've seen before. I just want to know how to proceed, you don't have to solve it for me :)

Homework Statement

Study the continuity of the following functions, defined by:

1- f(x) = lim (n^x-n^-x)/(n^x+n^-x) x∈|R
n->+∞

If you divide both numerator and denominator by n^x, you get
\frac{1- n^{-2x}}{1+ n^{-2x}}
Now suppose x> 0 and look at three cases, 0< x< 1, x= 1, x> 1.

Then divide both numerator and denominator by n^{-x} to get
\frac{n^{2x}- 1}{n^{2x}+ 1}
And do similary for x< 0.

2- f(x) = lim [ln(e^n+x^n)]/n x∈|R
n->+∞

The Attempt at a Solution

A function is continuos if its limit L exists and it equals f(L).
But the limit here is to +∞!
So, after computing the two limits for the given n->+∞, how do I go on studying the finction?

Many thanksss