limit of (fraction + fraction)

[idllotsaroms]

Hello, I've been stuck on this problem for quite some time.




Limit as x approaches 1: \(\displaystyle \displaystyle\frac{1}{x-1} + \frac{1}{x^2 - 3x + 2}\)

I tried to factor the \(\displaystyle \displaystyle\frac{1}{x^2 - 3x + 2}\) to (x-2)(x-1) so that I could multiply \(\displaystyle \displaystyle\frac{1}{x-1}\) top and bottom by (x-2) to have the common denominator \(\displaystyle \displaystyle\frac{(x-2)+1}{(x-1)(1-2)}\)
But even after this I do not believe I am getting the correct answer.

Any help is appreciated!

 

[Deleted member 4993]

Hello, I've been stuck on this problem for quite some time.




Limit as x approaches 1: \(\displaystyle \displaystyle\frac{1}{x-1} + \frac{1}{x^2 - 3x + 3}\)

I tried to factor the \(\displaystyle \displaystyle\frac{1}{x^2 - 3x + 3}\) to (x-2)(x-1) so that I could multiply \(\displaystyle \displaystyle\frac{1}{x-1}\) top and bottom by (x-2) to have the common denominator \(\displaystyle \displaystyle\frac{(x-2)+1}{(x-1)(1-2)}\)
But even after this I do not believe I am getting the correct answer.

Any help is appreciated!

As posted, the two sided limit of the given function does not exist.

 

[HallsofIvy]

You have \(\displaystyle \frac{1}{x- 1}+ \frac{1}{x^2- 3x+ 3}\). However, \(\displaystyle (x- 1)(x- 2)= x^2- 3x+2\), not \(\displaystyle x^2- 3x+ 3\). In fact, by the quadratic formula, \(\displaystyle x^2- 3x+ 3= 0\) has roots \(\displaystyle \frac{3\pm\sqrt{9- 12}}{2}\) which are complex numbers. \(\displaystyle x^2- 3x+ 3\) cannot be factored with real number coefficients.

 

[idllotsaroms]

Sorry, I've made a typo

The problem was \(\displaystyle \displaystyle\frac{1}{x-1} + \frac{1}{x^2 - 3x + 2}\), which I was factoring for. Wolfram says the answer I should be obtaining is -1, but I'm getting a 0/0

 

[srmichael]

Sorry, I've made a typo

The problem was \(\displaystyle \displaystyle\frac{1}{x-1} + \frac{1}{x^2 - 3x + 2}\), which I was factoring for. Wolfram says the answer I should be obtaining is -1, but I'm getting a 0/0

Now that you made the correction, go ahead and add these fractions together and simplfy. What do you get?

 

[idllotsaroms]

Now that you made the correction, go ahead and add these fractions together and simplfy. What do you get?

I got the answer as: \(\displaystyle \displaystyle\frac{(x-2)+1}{(x-1)(1-2)}\) because I cannot cancel our the (x-2) from the (1-2) in the denominator due to the numerator being addition (is that correct)? Which if i plug in 1 for x I get the answer as 1-2+1 =0 over (1-1)(1-2) = 0 or 0/0.

 

[srmichael]

I got the answer as: \(\displaystyle \displaystyle\frac{(x-2)+1}{(x-1)(1-2)}\) because I cannot cancel our the (x-2) from the (1-2) in the denominator due to the numerator being addition (is that correct)? Which if i plug in 1 for x I get the answer as 1-2+1 =0 over (1-1)(1-2) = 0 or 0/0.

Before you apply the limit as x approaches 1, further simplify....

\(\displaystyle \displaystyle\frac{(x-2)+1}{(x-1)(x-2)}\)

\(\displaystyle =\displaystyle\frac{x-1}{(x-1)(x-2)}\)

\(\displaystyle =\displaystyle\frac{1}{x-2}\)

Now apply the limit as x approaches 1.

 

[idllotsaroms]

my math skills are terrible....
Thanks srmichael!