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

If $t o 7^+$, the worth of $t - 7$ will be always positive, because if $t$ is pull close $7$ indigenous the ideal side, $t$ will constantly be a small bigger 보다 $7$, and also thus $t - 7$ will constantly be a small bigger 보다 $0$ (i.e., positive).

You are watching: Sqrt(7/3)

So simple the limit:

$lim limits_t o 7^+ dfracsqrt(t-7)^3t-7$

$lim limits_t o 7^+ dfracsqrt(t-7)^2sqrt(t-7)t-7$

$lim limits_t o 7^+ dfracsqrtt-7t-7$

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

$lim limits_t o 7^+ sqrtt-7= 0$

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answered Sep 4 "15 at 16:35

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