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