Question #c86dc

2 Answers
Jan 6, 2017

a=2

Explanation:

Using L'Hopital's rule:

lim_(x->oo)((ax+1)/(ax))^x = lim_(x->oo)e^ln(((ax+1)/(ax))^x)

=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 f(x)=e^x

Focusing on the new limit:

lim_(x->oo)xln((ax+1)/(ax)) = lim_(x->oo)ln(1+1/(ax))/(1/x)

Direct substitution leads to a 0/0 indeterminate form, and so we can apply L'Hopital's rule.

=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

lim_(x->oo)((ax+1)/(ax))^x=e^(lim_(x->oo)xln((ax+1)/(ax)))

=e^(1/a)

Setting this equal to sqrt(e) = e^(1/2), we get

e^(1/a) = e^(1/2)

=> ln(e^(1/a))=ln(e^(1/2))

=>1/a=1/2

:.a=2

Jan 6, 2017

a=2

Explanation:

An alternative answer, using algebra and the possible definition of e as e=lim_(x->oo)(1+1/x)^x:

lim_(x->oo)((ax+1)/(ax))^x = lim_(x->oo)(1+1/(ax))^x

=lim_(x->oo)((1+1/(ax))^(ax))^(1/a)

Let u = ax, and note that u->oo as x->oo for any a>0 (note that this assumes a>0).

=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 f(x) = x^c, c in RR

=e^(1/a)

Setting this result equal to sqrt(e) = e^(1/2), we get

e^(1/a) = e^(1/2)

:. a = 2