# Thread: Calculating the Limit of a Function!

1. ## 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

2. 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

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

4. Oh my gosh! The Mathematica is awesome!!!

5. thank you for letting me know the mathematica!! wow! I am getting very excited now XD

6. 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?

8. 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!

9. Originally Posted by Archie Meade
$\displaystyle \displaystyle\frac{4-5x}{1-x}}$

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 ...

10. 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

11. 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.

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

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 $\displaystyle y\rightarrow\infty$

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

13. Originally Posted by Archie 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 $\displaystyle y\rightarrow\infty$

$\displaystyle \displaystyle\lim_{y\rightarrow\ -\infty}e^y=0$
Yes, I see your logic here and perhaps I have complicated the matter Thanks