Solving for Parameter a in a Piecewise Function

[azatkgz]

Homework Statement

Find all values of the parameter a>0 such that the function

f(x)=\left\{\begin{array}{cc}\frac{a^x+a^{-x}-2}{x^2},x>0\\3ln(a-x)-2,x\leq0\end{array}\right

The Attempt at a Solution

\lim_{x\rightarrow 0}\frac{a^x+a^{-x}-2}{x^2}=0

0=3ln(a-0)-2\rightarrow a=e^{2/3}

 

[arildno]

Why should the limit value be 0??

All that is required is that AT x=0, both expressions should attain the same value!

Now, consider the function's expression for the non-positives.
Clearly, its limit at x=0 is 3ln(a)-2.
Thus, you are to determine those values of "a" so that the limiting value of the function expression for the positives equals 3ln(a)-2 as well.

 

[azatkgz]

ok
\lim_{x\rightarrow 0}\frac{a^x+a^{-x}-2}{x^2}=\frac{ln^2a}{2}
by L'Hopital Rule

\frac{ln^2a}{2}=3ln(a)-2

ln^2a-4lna+4=0

 

[arildno]

The limit of the upper expression should be \ln^{2}(a)

Thus, you have the quadratic in ln(a) to solve:
\ln^{2}(a)=3\ln(a)-2