Evaluating $\lim_{x\to1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$ I don't know how to solve these kind of limit problems. I was wondering if someone could help me about it to learn them. $$\lim_{x\to 1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$$ Thanks in advance.
 A: Hint
You can always write $f(x)-g(x)$ as $$\frac{f(x)^2-g(x)^2}{f(x)+g(x)}.$$
So, taking $$f(x)=\sqrt{x+1}+\sqrt{x^2+1}\quad \text{and}\quad g(x)=\sqrt{x^3+1},$$ at the numerator and $$f(x)=\sqrt{x-1}+\sqrt{x^2+1}\quad \text{and}\quad g(x)=\sqrt{x^4+1},$$
at the denominator should do the work.
A: Write $x=1+t$, and use $\sqrt{A+Bt}=\sqrt{A}+\tfrac{B}{2\sqrt{A}}t+o(t)$ for $A\ne0$ to write the limit as$$\lim_{t\to0^+}\frac{\sqrt{2}(1+\tfrac12o(t))+\sqrt{2t}(1+t)^{1/2}-\sqrt{2}(1+\tfrac34t+o(t))}{\sqrt{t}+\sqrt{2}(1+\tfrac12t+o(t))-\sqrt{2}(1+t+o(t))}.$$While the constant terms in the numerator and denominator cancel, the $\sqrt{t}$ terms don't, so the limit is$$\lim_{t\to0}\frac{\sqrt{2t}}{\sqrt{t}}=\sqrt{2}.$$
A: We can rewrite the numerator as $$N=\sqrt{x-1}\sqrt{x+1}+\frac{x(1-x)(1+x)}{\sqrt{1+x}+\sqrt{1+x^3}}=\sqrt{x-1}\left(\sqrt{1+x}-\frac{x(1+x)\sqrt{x-1}}{\sqrt{1+x}+\sqrt {1+x^3}}\right)$$ and hence $N/\sqrt {x-1}\to\sqrt{2}$. Similarly one can show that if $D$ is the denominator then $D/\sqrt{x-1}\to 1$. And therefore $N/D\to\sqrt {2}$.
Most algebraic limits can be handled using simple algebraic manipulation combined with the standard algebraic limit $\lim\limits _{x\to a} \dfrac{x^n-a^n} {x-a} =na^{n-1}$.
