how to find the limit of $\lim_{x \to0}\frac{\sqrt{x^2+x+1}-\sqrt{x+1}}{x^2}$

How to find the limit of $$\lim_{x \to0}\frac{\sqrt{x^2+x+1}-\sqrt{x+1}}{x^2}\,?$$ I tried L'Hospital's rule, but it didn't work well.

Can I have some assistance? Thank you in advance

Multiply numerator and denominator by $$\sqrt{x^2+x+1}+\sqrt{x+1}.$$ You will get $$\frac{1}{\sqrt{x^2+x+1}+\sqrt{x+1}}$$ and the limit is $$\frac{1}{2}$$

Since, near $$0$$,$$\sqrt{x^2+x+1}=1+\frac x2+\frac{3x^2}8+O(x^3)$$and$$\sqrt{x+1}=1+\frac x2-\frac{x^2}8+O(x^3),$$then$$\lim_{x\to0}\frac{\sqrt{x^2+x+1}-\sqrt{x+1}}{x^2}=\frac38-\left(-\frac18\right)=\frac12.$$

why apply l hopital when you can rationlaise..

$$\lim_{x \to0}\frac{\sqrt{x^2+x+1}-\sqrt{x+1}}{x^2}$$

and you will get the following

$$\lim_{x \to0}\frac{1}{\sqrt{x^2+x+1}+\sqrt{x+1}}$$

put $$x =0$$ you will get $$\frac{1}{2}$$

• Could you please explain how your answer is different from Dr Sonnhard's answer? – Toby Mak Apr 27 at 13:28
• i was wring the the answer and decorating it with latex...then when posted it...and aldready saw that someone posted the same solution – Marvel Maharrnab Apr 27 at 13:36
• That's fine, just checking. – Toby Mak Apr 27 at 13:37
• should i remove it? – Marvel Maharrnab Apr 27 at 13:38
• after posting ..i was thinking of approaching with another solution...and was quite impressed with jose carlos solution – Marvel Maharrnab Apr 27 at 13:44