Is it valid to simplify $\sqrt{(t-7)^3}$ to $(t-7)^2$?

Question

$\lim\limits_{t \to 7^+} \dfrac{\sqrt{(t-7)^3}}{t - 7}$

Can you simplify the top of the quotient to   $(t-7)^2$ ?

Or a more general question, does the square root of exponent 3 simplify to exponent 2?

Answer 1

$\sqrt{(t-7)^3}=(t-7)^{\frac{3}{2}}=(t-7)(t-7)^{\frac{1}{2}}$.

Answer 2

i would say $\sqrt{(x-7)^3}=(x-7)\sqrt{x-7}$ for $x>7$

Answer 3

If $t \to 7^+$, the value of $t - 7$ will be always positive, since if $t$ is approaching $7$ from the right side, $t$ will always be a little bigger than $7$, and thus $t - 7$ will always be a little bigger than $0$ (i.e., positive).

So simplifying the limit:

$\lim \limits_{t \to 7^+} \dfrac{\sqrt{(t-7)^3}}{t-7}$

$\lim \limits_{t \to 7^+} \dfrac{\sqrt{(t-7)^2}\sqrt{(t-7)}}{t-7}$

$\lim \limits_{t \to 7^+} \dfrac{|t-7|\sqrt{t-7}}{t-7}$

Since $t-7$ will always be positive, $|t-7|=t-7$, and they cancel out from the numerator and the denominator, leaving:

$\lim \limits_{t \to 7^+} \sqrt{t-7}= 0$