I was trying to find asymptotes for a function f ( x ) = x e <mrow class="MJ

Winigefx 2022-06-30 Answered
I was trying to find asymptotes for a function f ( x ) = x e 1 x 2
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

luisjoseblash2
Answered 2022-07-01 Author has 16 answers
When x is large, let
1 x 2 = t x = 2 + 1 t x e 1 x 2 = ( 2 t + 1 ) t e t
Now, using Taylor around t=0
( 2 t + 1 ) t e t = 1 t + 3 + 5 t 2 + 7 t 2 6 + O ( t 3 )
Back to x
x e 1 x 2 = 6 x 3 18 x 2 + 15 x + 1 6 ( x 2 ) 2
Long division
x e 1 x 2 = x + 1 + 5 2 x + O ( 1 x 2 )
which gives not only the slant asymptote but also shows how the function does approach it.
Did you like this example?
Subscribe for all access
Layla Velazquez
Answered 2022-07-02 Author has 11 answers
We have:
lim x + f ( x ) = lim x + x e 1 x 2 = +
We notice that e 1 x 2 1, so it's true the following asymptotic relation:
f ( x ) x
when x + . This is a necessary but not sufficient condition for the existence of the asympot.
We are searching for line of the type y = x + k , k R
lim x + f ( x ) x = lim x + x ( e 1 x 2 1 ) lim x + x 1 x 2 = 1
Here, we have used a very important asymptotic relation:
e f ( x ) 1 f ( x ) x x 0
With a function f ( x ) such that f ( x ) 0 when x x 0 R ¯
Here, I put f ( x ) = 1 x 2 that is such that f ( x ) 0 when x +
In conclusion, the asympot is:
y = x + 1
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions