Evaluating limit by factorization

[Joel Jacon]

Can anyone tell me how to solve the following limit by factorization method
$\lim{{x}\to{5}} \frac{x^3 + 3x^2 - 6x + 2}{ x^3 + 3x^2 - 3x - 1}$?Please tell me how to factorize such big equation?

 

[Siron]

Why do you want to factorize it?
The factorization method is useful when the limit is of an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. But this is not the case thus you can just plug in the value $x=5$.

 

[Joel Jacon]

But the answer given in my book is -11. While using direct substitution I get 9. How can you get -11

 

[Greg]

$$\lim_{x\to5}\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}=\frac{172}{184}=\frac{43}{46}$$$$\text{ }$$Are you sure you typed the problem correctly?

 

[Joel Jacon]

Yes, the question is correct. See the question 1 in the image

 

[MarkFL]

After saving the image, and rotating it so that it is not upside down, then straining my eyes to read the out of focus image, what I see is:

1.) \(\displaystyle \lim_{x\to5}\frac{2x^2+9x-5}{x+5}\)

Now, you can factor as follows (although it is not necessary):

\(\displaystyle \lim_{x\to5}\frac{(2x-1)(x+5)}{x+5}=\lim_{x\to5}2x-1=2(5)-1=9\)

Apparently what was meant, if an answer of $-11$ was given is:

\(\displaystyle \lim_{x\to-5}\frac{2x^2+9x-5}{x+5}=2(-5)-1=-11\)