Did I solve these limits right? I have some limits that I tried solving:
$\lim_{x\rightarrow2}(4x^5+2x^3-1)=4(2)^5+2(2)^3-1=143$
$\lim_{x\rightarrow1}(\frac{5x}{x-1})=\frac{5}{0}=does~not~exist$
$\lim_{x\rightarrow(-3)}(\frac{x^2-5x-24}{4x+12})=\frac{(x+3)(x-8)}{4(x+3)}=\frac{x-8}{4}=-\frac{11}{4}$
$\lim_{x\rightarrow4}(\frac{x^2-16}{x-4})=\frac{(x-4)(x+4)}{x-4}=x+4=8$
$\lim_{x\rightarrow0}(\frac{\sqrt{x+1}-1}{x})=\frac{\sqrt{x+1}-1}{x}*\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\frac{1}{\sqrt{x+1}+1}=\frac{1}{2}$
$\lim_{x\rightarrow27}(\frac{\sqrt[3]{x}-3}{x-27})=\lim_{u\rightarrow3}\frac{u-3}{u^3-27}=\frac{u-3}{(u-3)(u^2+3u+9)}=\frac{1}{u^2+3u+9)}=\frac{1}{27}$
Did I do everything right? Was there any mistakes in my approach?

Your approach to these is spot on, in all cases. You've done very well! Nice work!
Note, that when canceling common factors in the numerator and denominator, for example, say, the factor $(x-a)$ in a limit for which $x \to a$, you should specify $x \neq a$.
