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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

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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 $$\frac{1}{2}$$

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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$$\lim_{x\to0}\frac{\sqrt{x^2+x+1}-\sqrt{x+1}}{x^2}=\frac38-\left(-\frac18\right)=\frac12.$$

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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 $\frac{1}{2}$

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

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