# 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 [closed]

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.

• 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. Commented Feb 27, 2021 at 12:32
• Have you ever seen a problem to "rationalise the surds", e.g. $\frac{1}{\sqrt{1+x}-\sqrt{1-x}}$?
– user700480
Commented Feb 27, 2021 at 12:44
• math.stackexchange.com/q/2436856/42969 Commented Feb 27, 2021 at 12:52

Hint: put $$\;a=\sqrt[3]{1+x}\;,\;\;b=\sqrt[3]{1-x}\;$$ , then your eexpression is is
$$\frac{a^{3/2}-b^{3/2}}{a-b}=\frac{(a^{1/2}-b^{1/2})(a+a^{1/2}b^{1/2}+b)}{a-b}=\frac{a+a^{1/2}b^{1/2}+b}{a^{1/2}+b^{1/2}}$$