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 :)

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- Oct 4th 2010, 11:12 AMReal9999Calculating 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 AMmr fantastic
- Oct 4th 2010, 12:19 PMReal9999
Oh sorry about the title and thanks for your help.

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 PMReal9999
Oh my gosh! The Mathematica is awesome!!!

- Oct 4th 2010, 12:26 PMReal9999
thank you for letting me know the mathematica!! wow! I am getting very excited now XD

- Oct 4th 2010, 12:41 PMArchie Meade
- Oct 4th 2010, 12:45 PMReal9999
Ok! I have just purchased a copy! and downloading now! thank you so much!!!

- Oct 4th 2010, 12:51 PMReal9999
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 PMReal9999
- Oct 4th 2010, 01:07 PMReal9999
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 PMReal9999
- Oct 4th 2010, 02:01 PMArchie Meade
$\displaystyle \displaystyle\frac{4-5x}{1-x}$

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 AMReal9999