How can i find this limit without using L'Hôpital (or anything using derivatives for that matter)?
$$\lim_{x \to -8} \frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} $$
I have tried various substitutions and multiplying with conjugates etc, to no avail. I am probably doing something wrong, but I would really appreciate a good solution so I can follow each step and understand how to solve problems like this.
Answer
Substituting $$\sqrt[3]{x}=a$$ so we get
$$\frac{\sqrt{1-a^3}-3}{2+a}=\frac{1-a^3-9}{(2+a)(\sqrt{1-a^3}+3)}$$
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