How do you show that the limit as x approaches negative infinity of ex|x|a = 0?
a > 0
Well in a previous problem they asked me to show that a) limx->0 xaln(x) = 0 (a > 0)
That was pretty straight forward using L'hopital's rule on limx->0 ln(x) / x-a
But I'm having troubles with the followup question b) limx->- inf |x|aex = 0
They say: by putting x = ln (t) in a), show b)
We have the chain rule for limits, which says that for , if
- , and either
- g does not take the value d near c
then . In fact, c, d and e can also have the form or for ; in this case, the rule refers to the corresponding one-sided limit.
We need to prove where a > 0. Let b = 1 / a.
We know that where . Let . Then and for all x. By the chain rule, and . Therefore, .
Next, let . Then . By the chain rule again, and .