$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $

Question

Calculate below limit

$$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $$

Using L'Hôpital's rule might be too tedious. I wonder if there is a trick given the resemblence of numerator and denominator?

Answer 1

$$ \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} }=\frac{x+1-\frac{x^3}{ \sqrt{x^3+1} +\sqrt{x^4+1}}}{1-\frac{x}{ \sqrt{x+1}+\sqrt{x^2+1}}}\rightarrow\frac{2-\frac{1}{2\sqrt2}}{1-\frac{1}{2\sqrt2}}=$$ $$=\frac{4\sqrt2-1}{2\sqrt2-1}=\frac{15+2\sqrt2}{7}.$$

Answer 2

Using binomial first order expansion we have

• $\sqrt{x^3+1}=\sqrt{2+(x^3-1)}=\sqrt 2\sqrt{1+\frac{x^3-1}2}\sim \sqrt 2+\sqrt 2\frac{x^3-1}4$

• $\sqrt{x^4+1}\sim \sqrt 2+\sqrt 2\frac{x^4-1}4$

• $\sqrt{x+1}\sim \sqrt 2+\sqrt 2\frac{x-1}4$

• $\sqrt{x^2+1}\sim \sqrt 2+\sqrt 2\frac{x^2-1}4$

then

$$\frac{x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1}}{x-1 + \sqrt{x+1} - \sqrt{x^2+1}}\sim \frac{x^2-1+\sqrt 2\frac{x^3-1}4-\sqrt 2\frac{x^4-1}4}{x-1+\sqrt 2\frac{x-1}4-\sqrt 2\frac{x^2-1}4}=$$

$$=\frac{x+1+\sqrt 2\frac{x^2+x+1}4-\sqrt 2\frac{x^3+x^2+x+1}4}{1+\sqrt 2\frac{1}4-\sqrt 2\frac{x+1}4}\to \frac{2-\frac14 \sqrt 2}{1-\frac14 \sqrt 2}=\frac{15+2\sqrt2}{7}$$

Answer 3

$$L=\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} }$$ With L'Hospital's rule: $$L=\lim_{x \rightarrow 1} \frac{ 2x + 3/2x^2(x^3+1)^{-1/2} - 2x^3(x^4+1)^{-1/2} }{ 1 + 1/2(x+1)^{-1/2} -x (x^2+1)^{-1/2} }$$ $$L=\frac{ 2 + 3/2(2)^{-1/2} - 2(2)^{-1/2} }{ 1 + 1/2(2)^{-1/2} - (2)^{-1/2} }$$ $$L=\frac{ 2\sqrt 2 + 3/2 - 2 }{ \sqrt 2 + 1/2 -1 }$$ $$ \implies L=\frac{ 4\sqrt 2 -1 }{ 2\sqrt 2 - 1 }=\frac{ 15+2\sqrt 2 }{ 7 }$$