Limits

[TheBat]

I'm generally decent at limits but these two problems have been bugging me and I can't seem to get them right.

The first one is lim x->2 [(x^3) + 8]/[(x^4) - 16] = [(x+2)(x^2-2x+4)]/[(x^4)-(2^4)]

I can't figure how to factor out (x^4)-16 so that it doesn't divide by zero. The same thing happened later with me being unable to factor (x-1) out of (x^5)-1.

The second problem is lim x-> 2 [(1/x)-(1/2)]/(x-2) which I'm positively stumped by.

Any help would be appreciated. Sorry for the bad formatting but I couldn't figure out any better way to get it to work

 

[BigGlenntheHeavy]

First one:

\(\displaystyle \lim_{x\to2^{+}}\frac{x^{3}+8}{x^{4}-16} \ = \ \infty, \ \lim_{x\to2^{-}}\frac{x^{3}+8}{x^{4}-16} \ = \ -\infty, \ hence \ limit \ doesn't \ exist.\)

Second one:
\(\displaystyle \frac{[\frac{1}{x}-\frac{1}{2}]}{x-2} \ = \ \frac{2-x}{2x^{2}-4x}, \ see \ the \ Marqui \ on \ this \ one.\)

 

[TheBat]

Odd. The book lists the answer to the first one as -3/8 and the second as -1/4. Also the second is lim x->2 without 2 as a factor on the numerator. Is the book in the wrong? (it certainly has happened before)

 

[BigGlenntheHeavy]

Ok, I fix the second one, its limit is -1/4,

The first one is what it is.

Note: \(\displaystyle \lim_{x\to2} \ \frac{8-x^{3}}{x^{4}-16} \ = \ \frac{-3}{8}\)

 

[galactus]

Factor the top and bottom and you can see it. We have the difference of two cubes and the difference of two squares.

\(\displaystyle \lim_{x\to 2}\frac{-(x-2)(4+2x+x^{2})}{(x^{2}+4)(x+2)(x-2)}=\lim_{x\to 2}\frac{-(4+2x+x^{2})}{(x^{2}+4)(x+2)}\)Plug in x=2 and we see we get \(\displaystyle \frac{-12}{32}=\frac{-3}{8}\)