Evaluating limits

[ssb]

lim (x^5)-32/ x-2
x->2

Where do I start?
No clue...

 

[Ishuda]

lim (x^5)-32/ x-2
x->2

Where do I start?
No clue...

What is (x-2) * (x⁴ + 2x³ + 4x² + 8x + 16)?

 

[Deleted member 4993]

lim [(x^5)-32]/ [x-2]
x->2

Where do I start?
No clue...

Those [] are very important - otherwise your problem need to be interpreted differently [i.e. different from Ishuda's interpretation]

 

[Steven G]

lim (x^5)-32/ x-2
x->2

Where do I start?
No clue...

You need to recall what you learned in pre calculus and how to use bracket. Since x=2 makes your numerator and denominator equal to 0 then we know that the the numerator and denominator each have a factor of (x-2). The denominator is only (x-2) so it can't be factored any more but the numerator is more than (x-2). Again, we know one factor and need to find the 2nd factor (which may or may not be factorable itself). You can either divide (x^5 - 32) by (x-2) using long or synthetic division to find that (x^5- 32) = (x-2)(p(x). Then lim as x goes to 2 of [(x-2)p(x)/(x-2)] = lim as x goes to 2 of p(x) = p(2). So you must find this p(x)!