Consider the following limit (Exercise 1.10.3 (b) Problems in Calculus of One Variable by I.A. Maron):
$$\lim_{x\to-1} \frac{x+1}{(x+17)^{1/4}-2}$$
To solve this limit we substitute $x = z^4-17$ . As $x\to-1$ , we get that $z$ tends to two values! i.e. $z\to\pm2$.
This gives $$\lim_{z\to\pm2}(z+2)(z^2+4)$$ Which gives $32$ when $z\to2$ and $0$ when $z\to-2$.
Graphing the original function, we get that $32$ (This is the answer given in the book).
The problem can be prevented by substituting $z=(x+17)^{1/4}$ which gives $z\to2$ as $x\to-1$.
Here are my questions:
- What does the extra root mean/signify?
- How would I know (without referring to the answer or graph) which root is to be taken?
- From what I understand, this issue arises because we have an even radical (fourth root) we have to get rid of by substitution . What should be my approach in such problems?
Even roots can only return positive values. In this case $$(x+17)^{1/4} \to 16^{1/4} = \big( (\pm 2)^4 \big)^{1/4} = | \pm 2 | = 2$$ as $x \to -1$.