I define "standardized beta" as the slope of a regression line when all variables (all $X$'s and $Y$) have been standardized first. If you have a simple linear regression model (i.e., only one $X$ variable), the standardized beta is the same as Pearson's product-moment correlation, $r$1. As a result, the standardized beta is bound by the interval $[-1,\ 1]$, just like $r$ is. The reason is given by @whuber in his comment above. However, it might help to try to work through this more slowly.
Let's start by considering the formulas for the estimated slope of a regression line, $\hat\beta_1$ and for $r$:
$$
\hat\beta_1=\frac{\text{Cov}(x,y)}{\text{Var}(x)} \qquad\qquad r=\frac{\text{Cov}(x,y)}{\text{SD}(x)\text{SD}(y)}
$$
Now, if both $x$ and $y$ have been standardized first (so that their means are $0$ and their SDs are $1$), then the denominator of $\hat\beta_1$, i.e. the variance of $x$, will be $1^2=1$, and the denominator or $r$, i.e. the SD of $x$ times the SD of $y$, will be $1\times 1=1$. They will be the same. And the numerators are the same no matter what. Thus, the standardized beta is the same as $r$, and has all the same properties (e.g., the same possible range).
On an intuitive level, we can still ask the question why can't the standardized beta / $r$ go above 1? It seems like it ought to be possible to do this. Let's try to make an example. I'll use R. I don't know if you use R, but you can download it for free and run this example; I'll try to make it as self-explanatory as possible.
set.seed(4077) # this makes the example exactly reproducible
# here are the true parameters we'll use:
N = 30 # we will work with 30 data
b0 = 5 # the true intercept will be 5
b1 = 0 # at first, the true intercept is 0, no relationship
# let's make our X data & some residuals:
resids = rnorm(30, mean=0, sd=1)
x = rnorm(N, mean=50, sd=7)
# now we can generate Y from X, our residuals & our parameters:
y = b0 + b1*x + resids
# let's get the means & SDs of X & Y, & their covariance:
mean(x) # 51.72901
sd(x) # 7.7859
mean(y) # 4.82287
sd(y) # 0.8541943
cov(x,y) # 1.71654
# with these we can predict the estimated slope & correlation:
cov(x,y)/(sd(x)^2) # 0.02831629
cov(x,y)/(sd(x)*sd(y)) # 0.2581003
# let's check the estimated slope & correlation:
coef(lm(y~x))[2] # 0.02831629
cor(x,y) # 0.2581003
# what happens to the slope if we standardize both x & y first?
x.s = (x - mean(x))/sd(x)
y.s = (y - mean(y))/sd(y)
# the slope now equals the correlation above:
coef(lm(y.s~x.s))[2] # 0.2581003
In that case the true value of b1 was $0$, let's make it $1$:
b1 = 1
y1 = b0 + b1*x + resids
# let's see what happened:
mean(y1) # 56.55189
sd(y1) # 8.048787
cov(x,y1) # 62.33678
# calculating b1 & r:
cov(x,y1)/(sd(x)^2) # 1.028316
cov(x,y1)/(sd(x)*sd(y1)) # 0.9947298
# checking:
coef(lm(y1~x))[2] # 1.028316
cor(x,y1) # 0.9947298
# let's try the standardized version:
y1.s = (y1 - mean(y1))/sd(y1)
# here is the standardized beta (from now on, I'll dispense with
# also calculating r & then double checking with pre-set functions):
cov(x.s,y1.s)/(sd(x.s)^2) # 0.9947298
That looks about right, so let's make b1=2:
b1 = 2
y2 = b0 + b1*x + resids
# here's the estimated slope:
cov(x,y2)/(sd(x)^2) # 2.028316
# and now we can see the standardized beta:
y2.s = (y2 - mean(y2))/sd(y2)
cov(x.s,y2.s)/(sd(x.s)^2) # 0.9986374
What happened? The slope came out right, but the standardized beta didn't become larger than $1$. Let's try a bigger number, b1=7:
b1 = 7
y7 = b0 + b1*x + resids
# here's the estimated slope:
cov(x,y7)/(sd(x)^2) # 7.028316
# and now we can see the standardized beta:
y7.s = (y7 - mean(y7))/sd(y7)
cov(x.s,y7.s)/(sd(x.s)^2) # 0.9998863
We're still not getting standardized betas $>1$. The reason is that, although the unstandardized slope and the covariance is getting larger and larger, the standard deviation of $y$ is getting larger too.
cov(x,y2) # 122.957
cov(x,y7) # 426.0582
sd(y2) # 15.81382
sd(y7) # 54.72799
Once that fact is incorporated, the standardized beta is constrained to fall within the interval $[-1,\ 1]$.
1. For more on the relationship between correlation and regression, it may help you to read my answer here: What is the difference between linear regression on y with x and x with y?
