# Homework Help: Continuity of a Function.

1. Oct 21, 2007

### azatkgz

1. The problem statement, all variables and given/known data

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

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

3. The attempt at a solution

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

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

2. Oct 21, 2007

### 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.

3. Oct 21, 2007

### azatkgz

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

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

$$ln^2a-4lna+4=0$$

Last edited: Oct 21, 2007
4. Oct 21, 2007

### arildno

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

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