lim as x approaches 4 from the left, (x+4)/(x^2-16)

Question

anonymous · Answer

lim as x  approaches -4, (x+4)/(x^2-16) 
lim as x approaches 0, for the same function 
lim as x approaches infinity

abb0t · Answer

Take note of the denominator:

$$\lim_{x \rightarrow 4^-} \frac{ x+4 }{ x^2-16 } $$

$$x^2-16 = (x+4)(x-4)$$
 
cancel like terms. That should help guide you in the right direction.

anonymous · Answer

i have 1/0 as answer, does that mean the limit does not exist ?

abb0t · Answer

It's approaching from -∞

anonymous · Answer

f(x)= 1/(x-4)
thats the function i use for all the problems

abb0t · Answer

Yeah. But it's approaching 4 from the left, meaning negative. That's why it's -∞

anonymous · Answer

so for x approaching -4 i would get 1/(-4-4) which is 1/-8
how would i write the limit for that

anonymous · Answer

and x approaching 0 i would get 1/-4 how would i state the limit for that and as x approaches infinity i really don't get it