dusty wadi
"Find the limit. Use l'Hopital's rule where appropriate. If there is a more elementary method, try using it. If l'Hopital's rule doesn't apply, explain why."
The limit is lim (as x->∞) of the function (1+ (a/x))^(bx)
I know that for limits of one function divided by another, implementing l'Hopital's rule is pretty simple. I'm not sure where to use it here, however.
unkempt jacinth
I can tell you rn it’s 1
At least if b>0
The denominator grows a lot quicker than the numerator
As x gets big
Oh shit I misread it as $1+(\frac{a}{x})^{bx}$
raw fieldBOT
trigonometric identity
unkempt jacinth
Your limit is equal to $\exp(\lim_{x\to \infty}bx\ln(1+\frac{a}{x}))$ a.k.a. $\exp(\lim_{x\to \infty}\frac{\ln(1+\frac{a}{x})}{\frac{1}{bx}})$
raw fieldBOT
trigonometric identity
unkempt jacinth
Tada you now have a 0/0 form
But be careful and make sure to confirm that hopitals rule applies
Check that all the limits exist
They do btw
unkempt jacinth
Your limit is equal to $\exp(\lim_{x\to \infty}bx\ln(1+\frac{a}{x}))$ a.k.a. $\e...
This is using the composition rule for limits
unkempt jacinth
"Find the limit. Use l'Hopital's rule where appropriate. If there is a more elem...
As for there being a more elementary method, I feel like there’s is but idk what. That limit sure looks a lot like e
$e^x=\lim_{n\to \infty}(1+\frac{x}{n})^n$
raw fieldBOT
trigonometric identity
