# Testing extremities!

• Aug 18th 2009, 02:33 AM
Modernized
Testing extremities!
I have,
y= e^x / x

Now, I want to test the extremities, so I sub in
x -> +ve Infinity
so,
y -> 0

Right? However, when I graphed it on the calculator.. when x reached positive infinity, the y is not reaching zero.. it curved up to positive infinity..

Can anyone explain this? Thanks!
• Aug 18th 2009, 05:08 AM
stapel
Quote:

Originally Posted by Modernized
I have y= e^x / x

Does the above mean the following?

. . . . . $y\, =\, \frac{e^x}{x}$

Assuming so...

Quote:

Originally Posted by Modernized
I want to test the extremities...

Does the above means "find the horizontal asymptote" or "find the limit as x gets arbitrarily large" or something similar?

Assuming so...

Quote:

Originally Posted by Modernized
...so I sub in
x -> +ve Infinity

Does the above mean "So I take the limit of the expression, as x tends toward positive infinity"?

Assuming so...

Quote:

Originally Posted by Modernized
so,
y -> 0

How did you arrive at this limit value, seeing that the exponential numerator grows much more quickly than does the linear denominator?

Kindly please reply with corrections, confirmations, and / or clarification. Thank you! (Wink)
• Aug 18th 2009, 05:33 AM
Modernized
Yes, all of your assumptions are correct! Sorry for the doggy worded question!

"How did you arrive at this limit value, seeing that the exponential numerator grows much more quickly than does the linear denominator?"

y=e^x / x
as x approached +ve infinity, the numerator will definitely grow faster than the denominator

But! my teacher told me that as x -> + infinity,
y will get smaller and smaller, hence approaches 0.. is this correct?
• Aug 18th 2009, 06:43 AM
mr fantastic
Quote:

Originally Posted by Modernized
Yes, all of your assumptions are correct! Sorry for the doggy worded question!

"How did you arrive at this limit value, seeing that the exponential numerator grows much more quickly than does the linear denominator?"

y=e^x / x
as x approached +ve infinity, the numerator will definitely grow faster than the denominator

But! my teacher told me that as x -> + infinity,
y will get smaller and smaller, hence approaches 0.. is this correct?

It's more likely that your teacher told you that as $x \rightarrow {\color{red}-} \infty$ then $y \rightarrow 0$. Either that, or the function is actually $y = \frac{e^{{\color{red}-} x}}{x}$.
• Aug 18th 2009, 07:45 AM
nikhil
Quote:

Originally Posted by Modernized
I have,
y= e^x / x

Now, I want to test the extremities, so I sub in
x -> +ve Infinity
so,
y -> 0

Right? However, when I graphed it on the calculator.. when x reached positive infinity, the y is not reaching zero.. it curved up to positive infinity..

Can anyone explain this? Thanks!

just find limit when x approaches to +infinity
L'Hospital's rule may be used which straightly gives limit to be +infinity