Continuity of Functions with Limits to Infinity

In summary: Hi Felafel! :smile: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! :wink:Hi Felafel! :smile:
  • #1
Felafel
171
0
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
 
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  • #2
Hi Felafel! :smile:
Felafel said:
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! :wink:
 
  • #3
tiny-tim said:
Hi Felafel! :smile:


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! :wink:

thank you :)!
just.. random values?
 
  • #4
Felafel said:
just.. random values?

yup! :smile:

usually works! :biggrin:
 
  • #5
Felafel said:
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 [itex]n^x[/itex], you get
[tex]\frac{1- n^{-2x}}{1+ n^{-2x}}[/tex]
Now suppose x> 0 and look at three cases, 0< x< 1, x= 1, x> 1.

Then divide both numerator and denominator by [itex]n^{-x}[/itex] to get
[tex]\frac{n^{2x}- 1}{n^{2x}+ 1}[/tex]
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
 

1. What is continuity of functions?

Continuity of functions is a fundamental concept in calculus and real analysis that describes the smoothness and unbroken nature of a function. A function is said to be continuous if it has no sudden jumps, breaks, or holes in its graph.

2. How is continuity of functions defined mathematically?

A function f is continuous at a point x = a if the limit of f(x) as x approaches a is equal to the value of f(a). This can be written as: lim x→a f(x) = f(a). In other words, the function has the same output at a point as it does when approaching that point from both the left and right sides.

3. What is the difference between pointwise continuity and uniform continuity?

Pointwise continuity means that a function is continuous at every individual point in its domain. Uniform continuity, on the other hand, means that the function is continuous over an entire interval or set, rather than just at specific points.

4. Can a function be continuous at a point but not on an interval?

Yes, it is possible for a function to be continuous at a specific point but not on an interval. This occurs when the function has a discontinuity (such as a jump or infinite limit) at a different point within the interval.

5. How can we determine if a function is continuous using a graph?

A function is continuous on a closed interval if its graph is a continuous, unbroken curve with no gaps or holes. This means that the graph can be drawn without lifting the pen from the paper. If there are any sudden jumps or breaks in the graph, the function is not continuous on that interval.

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