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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.

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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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Thank you for the pointers, I had to work with it a bit before realising all the steps, but this is what I came up with, posting it here so it can help others.

Substituting $ \sqrt[3]{x} $ with $ t $.

Since $ x \to -8 $, $ t \to \sqrt[3]{-8} = -2 $ the limit is now rewritten as \begin{align*} \lim_{t \to -2} \frac{\sqrt{1-t^3}-3}{2 + t}. \end{align*} The fraction is multiplied with the conjugate for the numerator \begin{align*} \lim_{t \to -2} -\frac{t^3 + 8}{(2+t)(\sqrt{1-t} + 3)}. \end{align*} $ 2+t $ is factored out of the numerator and cancelled: \begin{align*} \lim_{t \to -2} - \frac{t^2 - 2t + 4}{\sqrt{1-t^3} + 3}. \end{align*} It is now possible to plug in the value for $ t $: \begin{align*} -\frac{(-2)^2 - 2(-2) + 4}{\sqrt{1-(-2)^3} + 3} = - \frac{12}{6} = -2. \end{align*}

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Let $2+\sqrt[3]x=y\implies x=(y-2)^3$

$$\lim_{y\to0}\dfrac{\sqrt{1-(y-2)^3}-3}y$$

$$=\lim_{y\to0}\dfrac{\sqrt{9-12y+6y^2-y^3}-3}y$$

$$=\lim_{y\to0}\dfrac{9-12y+6y^2-y^3-3^2}y\cdot\dfrac1{\lim_{y\to0}\sqrt{9-12y+6y^2-y^3}+3}=?$$

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