Question

Find the limit of the function algebraically.

Answer 1

@agreene

Answer 2

my answer; 8

Answer 3

@ganeshie8 help!

Answer 4

how 8 ?

Answer 5

i dont know what i was thinking, it is 0

Answer 6

i plug in 0, and got 0 after i solved it @ganeshie8

Answer 7

if you plug in 0 you should get \(\frac00\) which is indeterminate

Answer 8

so it would not exist?

Answer 9

it doesn't exist but thats not why.

Answer 10

Hate doing that... why not just use L'H'S, derive top and bottom.\[\color{darkgreen}{\lim_{x \rightarrow 0}~ \frac{x^2-2x}{x^3}} \]
\[\color{darkgreen}{\lim_{x \rightarrow 0}~ \frac{\frac{d}{dx}(x^2-2x)}{\frac{d}{dx}(x^3)}} \]
\[\color{darkgreen}{\lim_{x \rightarrow 0}~ \frac{2x-2}{3x^2}} \]and derive more...\[\color{darkgreen}{\lim_{x \rightarrow 0}~ \frac{\frac{d}{dx}(2x-2)}{\frac{d}{dx}(3x^2)}} \]
\[\color{darkgreen}{\lim_{x \rightarrow 0}~ \frac{2}{9x}} \]

Yes the limit does exist, or at least doesn't seem to exist. if it was x² on the bottom, it would have been there.

Answer 11

With whatever method it doesn't exist.

Answer 12

I'm a big fan of L'Hopital's but because the power in the bottom is larger than the top, you can assume it's going to give indeterminate results.

If you plug in a value really close to 0 on the left and right side of 0 you should find that
\[\lim_{x\rightarrow0^+}f(x)=-\infty\\ \lim_{x\rightarrow0^-}f(x)=\infty\]
because left and right are not the same the limit DNE