# Calculating the Limit of a Function!

• Oct 4th 2010, 11:12 AM
Real9999
Calculating the Limit of a Function!
Can someone show me the steps of calculating the limit of the following function?

When x approaches 1 from the left (1 minus), the limit of exp {(4-5x)/(1-x)}.

Thank you very much :)
• Oct 4th 2010, 11:50 AM
mr fantastic
Quote:

Originally Posted by Real9999
Can someone show me the steps of calculating the limit of the following function?

When x approaches 1 from the left (1 minus), the limit of exp {(4-5x)/(1-x)}.

Thank you very much :)

limit of Exp&#91;&#40;4-5x&#41;&#47;&#40;1-x&#41;&#93; as x approaches 1 - Wolfram|Alpha
• Oct 4th 2010, 12:19 PM
Real9999

But can you actually show me how to come to the 0? I know the answer is 0 but I am struggling in getting 0:(
• Oct 4th 2010, 12:24 PM
Real9999
Oh my gosh! The Mathematica is awesome!!!
• Oct 4th 2010, 12:26 PM
Real9999
thank you for letting me know the mathematica!! wow! I am getting very excited now XD
• Oct 4th 2010, 12:41 PM
Quote:

Originally Posted by Real9999

But can you actually show me how to come to the 0? I know the answer is 0 but I am struggling in getting 0:(

$\displaystyle \displaystyle\frac{4-5x}{1-x}}$

If x<1 and approaching 1, then the denominator is positive and approaching 0.

The numerator is approaching -1.

What then is the fraction approaching?
• Oct 4th 2010, 12:45 PM
Real9999
• Oct 4th 2010, 12:51 PM
Real9999
Oh, yes... what I was concerning was that since it is 1 - x in the denominator, I cannot directly apply the limit to the fraction.

But thanks Archie! I know how to work it out now! What I need is to take the inverse, (1/x) / (4 - 5x) and evaluate the limit when x approaches 1 from the left and then take the inverse again. That is, the inverse of the limit of the inverse of a fraction is equivalent to the limit of the original function!

I am so happy today! Thanks for letting me know the existence of mathematica!
• Oct 4th 2010, 12:59 PM
Real9999
Quote:

$\displaystyle \displaystyle\frac{4-5x}{1-x}}$

If x<1 and approaching 1, then the denominator is positive and approaching 0.

The numerator is approaching -1.

What then is the fraction approaching?

What you are saying here is to directly apply limit to the denominator? does that actually work?

I think my approach of taking inverse of the function is still flawed ...
• Oct 4th 2010, 01:07 PM
Real9999
Another thing concerning me is that is e^(4-5x)/(1-x) actually has a limit at or approaches 0? I know that exponential function is always positive and not defined at 0 :(
• Oct 4th 2010, 01:29 PM
Real9999
I think the answer giving by the mathematica is wrong... just simply by looking at the graph generated by mathematica, it does not look like an exponential function at all...>_< Please correct me if I am wrong... thank you very much.
• Oct 4th 2010, 02:01 PM
$\displaystyle \displaystyle\frac{4-5x}{1-x}$

The fraction is not defined for $\displaystyle x=1$ as this will cause the denominator to be zero.

$\displaystyle x=1$ does not cause a problem for the numerator, however.

As x approaches 1 from below, the fraction is negative, the denominator approaches zero and hence the fraction approaches $\displaystyle -\infty$

As x approaches 1 from above, the fraction is positive, the denominator approaches zero and hence the fraction approaches $\displaystyle +\infty$

$\displaystyle \displaystyle\lim_{y\rightarrow\infty}e^y$ doesnt exist since $\displaystyle e^y\rightarrow\infty$ as $\displaystyle y\rightarrow\infty$

$\displaystyle \displaystyle\lim_{y\rightarrow\ -\infty}e^y=0$
• Oct 5th 2010, 01:53 AM
Real9999
Quote:

$\displaystyle \displaystyle\frac{4-5x}{1-x}$

The fraction is not defined for $\displaystyle x=1$ as this will cause the denominator to be zero.

$\displaystyle x=1$ does not cause a problem for the numerator, however.

As x approaches 1 from below, the fraction is negative, the denominator approaches zero and hence the fraction approaches $\displaystyle -\infty$

As x approaches 1 from above, the fraction is positive, the denominator approaches zero and hence the fraction approaches $\displaystyle +\infty$

$\displaystyle \displaystyle\lim_{y\rightarrow\infty}e^y$ doesnt exist since $\displaystyle e^y\rightarrow\infty$ as $\displaystyle y\rightarrow\infty$

$\displaystyle \displaystyle\lim_{y\rightarrow\ -\infty}e^y=0$

Yes, I see your logic here and perhaps I have complicated the matter :( Thanks