Two calculus questions

Can someone check the answers I worked out to these two questions -

1). [latex] \lim_{x \to 0} \ ((x^3 \ - \ 5x)\cos \ (\frac {1}{x^8}))[/latex]

[latex]-1 \ $\leq$ \cos (\frac {1}{x^8}) \ $\leq$ 1[/latex]

Multiply across by [latex]x^3 \ - \ 5x[/latex]

[latex]-(x^3 \ - \ 5x) \ $\leq$ ((x^3 \ - \ 5x)\cos (\frac {1}{x^8}) \ $\leq$ (x^3 \ - \ 5x)[/latex]


The limits on each side are both 0
[latex] \lim_{x \to 0} \ -x^3 \ + \ 5x = 0[/latex]

[latex] \lim_{x \to 0} \ x^3 \ - \ 5x = 0[/latex]

Therefore, by the squeeze theorem [latex] \lim_{x \to 0} \ ((x^3 \ - \ 5x)\cos \ (\frac {1}{x^8})) = 0[/latex] too. Is that correct?


2. I'm given [latex] \lim_{x \to 4} \frac {f(x)}{x^2} = 3[/latex] and I have to calculate (a) [latex] \lim_{x \to 4} f(x)[/latex] and (b) [latex] \lim_{x \to 4} \frac {f(x)}{x}[/latex]

(a)
[latex] \lim_{x \to 4} f(x) [/latex] / [latex]\lim_{x \to 4} x^2[/latex] = 3 (Quotient rule)

[latex] \lim_{x \to 4} f(x) = 3 ( \lim_{x \to 4} x^2)[/latex]

[latex] \lim_{x \to 4} f(x) = 3 (16) = 48[/latex]


(b) [latex] \lim_{x \to 4} \frac {f(x)}{x}[/latex]

[latex] \lim_{x \to 4} f(x)[/latex] / [latex] \lim_{x \to 4} x[/latex]

= 48 / 4 = 12


How do these answers look, are they correct?