Question

\[f(x)= e ^{\frac{ 2x }{ x ^{2} + 1 }}\]
asymptotes

Answer 1

Asymptotes could be horizontal, vertical or slanted. The last category is harder to find, so we'll hope they do not occur ;).
For e^x we know there is a horizontal asymptote (x-axis), because:\[\lim_{x \rightarrow -\infty}e^x=0\]We just have to see if there is a way in which 2x/(x²+1) also goes to −∞, because e to that power will tend to 0.

Answer 2

i dont understand how to solve this question

Answer 3

You do know the function g(x)=e^x, though.
Its graph has the x-axis as asymptote, because the more negative the number you put in, the less e^x becomes: it tends to 0 as x becomes very large negative.
Do you understand this?

Answer 4

This is you "normal" e^x graph:

Answer 5

\[e ^{x} = 0\]

there is not horizontal asymptote

Answer 6

The exponent of f looks rather complicated, but we are lucky: the denominator is x²+1.
As x² is 0 or bigger, adding 1 guarantees that it is always greater than 0. So dividing by 0 cannot occur. So no asymptotes because of a zero value in the denominator.

Now all we have to do is see what happens with the exponent when x goes to plus or minus infinity:\[\lim_{x \rightarrow \infty} \frac{ 2x }{ x^2+1 }\]and\[\lim_{x \rightarrow -\infty}\frac{ 2x }{ x^2+1 }\]

It is easy to see that for very large x (pos or neg), the denominator always is much larger than the numerator, because it has a squared x. So these limits are 0.

Answer 7

when denominator is much larger than the numerator ..Always limit is 0

Answer 8

So for very large values (pos or neg) of x, the exponent of function f goes to 0, so f(x) tends to e^0 = 1. This means that y = 1 is horizontal asymptote. Graph:

Answer 9

You are right! and therefore e^0 = 1.
See? f(x) = e^(number that goes to 0, for x to infinity) = 1

Answer 10

ok i understand

Answer 11

Happy New Year!

Answer 12

same 2 u