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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1$\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$– Martin RCommented Feb 27, 2021 at 12:32
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$\begingroup$ Have you ever seen a problem to "rationalise the surds", e.g. $\frac{1}{\sqrt{1+x}-\sqrt{1-x}}$? $\endgroup$– user700480Commented Feb 27, 2021 at 12:44
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$\begingroup$ math.stackexchange.com/q/2436856/42969 $\endgroup$– Martin RCommented Feb 27, 2021 at 12:52
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1 Answer
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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, 2021 at 12:44