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
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityHow 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 $\frac{1}{2}$