Question #c86dc
2 Answers
Explanation:
Using L'Hopital's rule:
=lim_(x->oo)e^(xln((ax+1)/(ax))
=e^(lim_(x->oo)xln((ax+1)/(ax)))
The last equality follows from the continuity of the function
Focusing on the new limit:
Direct substitution leads to a
=lim_(x->oo)(d/dxln(1+1/(ax)))/(d/dx1/x)
=lim_(x->oo)(1/(1+1/(ax))(-1/(ax^2)))/(-1/x^2)
=lim_(x->oo)1/(a+1/x)
=1/a
Substituting this back into our original limit, we have
=e^(1/a)
Setting this equal to
Explanation:
An alternative answer, using algebra and the possible definition of
=lim_(x->oo)((1+1/(ax))^(ax))^(1/a)
Let
=lim_(u->oo)((1+1/u)^u)^(1/a)
=(lim_(u->oo)(1+1/u)^u)^(1/a)
The above equality follows from the continuity of the function
=e^(1/a)
Setting this result equal to