Find limit (without using L' Hospital Rule) I can find this limit using L' Hospital Rule, I do not know how to do it without that

Question

Find limit (don't use Lophital rule) $$\lim _{x\to 0}\left(\frac{\sqrt{1+x}\:-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}\:}\right)$$

I can find this limit using L' Hospital Rule, I do not know how to do it without that.

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Have you ever seen a problem to "rationalise the surds", e.g. $\frac{1}{\sqrt{1+x}-\sqrt{1-x}}$? — , Feb 27 '21 at 12:44

score 0 · Accepted Answer · answered Feb 27 '21 at 12:44

0

Hint: put $\;a=\sqrt[3]{1+x}\;,\;\;b=\sqrt[3]{1-x}\;$ , then your eexpression is is

$$\frac{a^{3/2}-b^{3/2}}{a-b}=\frac{(a^{1/2}-b^{1/2})(a+a^{1/2}b^{1/2}+b)}{a-b}=\frac{a+a^{1/2}b^{1/2}+b}{a^{1/2}+b^{1/2}}$$

answered Feb 27 '21 at 12:44

DonAntonio

1 Answers1