We are asked to find the value of the limit:

$\lim_{h \to 0} \frac{f(1-h) - f(1)}{h^3 + 3h}$

where $f(x) = 3x^{10} - 7x + 5x - 21x^3 + 3x^2 - 7$.

Step 1: Simplify $f(x)$

First, simplify the function $f(x)$:

$f(x) = 3x^{10} - 7x + 5x - 21x^3 + 3x^2 - 7$ $f(x) = 3x^{10} - 2x - 21x^3 + 3x^2 - 7$

Step 2: Calculate $f(1-h)$ and $f(1)$

Next, substitute $1-h$ into the function to find $f(1-h)$:

$f(1-h) = 3(1-h)^{10} - 2(1-h) - 21(1-h)^3 + 3(1-h)^2 - 7$

And for $f(1)$:

$f(1) = 3(1)^{10} - 2(1) - 21(1)^3 + 3(1)^2 - 7$ $f(1) = 3(1) - 2(1) - 21(1) + 3(1) - 7$ $f(1) = 3 - 2 - 21 + 3 - 7 = -24$

Step 3: Expand $f(1-h)$ for small $h$

Now, expand $f(1-h)$ using the binomial theorem:

$(1-h)^{10} = 1 - 10h + \frac{45h^2}{2} - \dots$ $(1-h)^3 = 1 - 3h + 3h^2 - h^3$ $(1-h)^2 = 1 - 2h + h^2$

Substituting these expansions into $f(1-h)$:

[ f(1-h