# Limit Question $\lim_{x\to\infty} \sqrt{x^2+1}-x+1$

I understand the answer is 1 which kind of makes sense intuitively but I can't seem to get there. I would appreciate if someone pointed out which line of my reasoning is wrong, thanks. I tried writing all my steps

$$$$\lim_{x\to\infty} \sqrt{x^2+1}-x+1$$$$

$$$$\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}$$$$

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

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

$$$$\lim_{x\to\infty} \frac{x \left( x - 1 +\frac{2}{x}\right)}{x \left( \sqrt{1+\frac{1}{x}}+1-\frac{1}{x} \right)}$$$$

$$$$\lim_{x\to\infty} \frac{x - 1 +\frac{2}{x}}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}$$$$

$$$$\lim_{x\to\infty} \frac{\infty - 1 + 0}{1+1-0}$$$$

$$$$\lim_{x\to\infty} \frac{\infty - 1}{2} = \infty$$$$

$$$$\lim_{x\to\infty} \sqrt{x^2+1}-x+1$$$$

$$$$\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}$$$$

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

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

$$$$\lim_{x\to\infty} \frac{x}{x} \frac{2}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}$$$$

$$$$\frac{2}{1+1} = 1$$$$

• As an addition you may want to consider this answer as well, in particular the second part of it – roman Dec 17 '18 at 11:38

From here we have

$$\frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}=\frac{(\sqrt{x^2+1})^2-(x-1)^2}{\sqrt{x^2+1}+(x-1)}=$$$$=\frac{x^2+1-x^2+2x-1}{\sqrt{x^2+1}+(x-1)}=\frac{2x}{\sqrt{x^2+1}+(x-1)}$$

• Ah... brainfart. Thanks. I'll accept as answer once the time limit is up. – Quaz Dec 17 '18 at 7:44
• @Quaz You are welcome bye! – user Dec 17 '18 at 7:46

At line $$3$$ you should have $$\frac{x^2+1-(x-1)^2}{\sqrt{x^2+1}+x-1}.$$

Set $$1/x=h\implies h\to0^+$$

and $$\sqrt{x^2+1}=\dfrac{\sqrt{1+h^2}}{|h|}=\dfrac{\sqrt{1+h^2}}h$$ as $$h>0$$ as $$h\to0^+$$

So, we have $$\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}$$

$$=1+\lim_{h\to0^+}\dfrac{\sqrt{1+h^2}-1}h$$

$$=1+\lim_{h\to0^+}\dfrac{1+h^2-1}h\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+h^2}+1}$$

$$=?$$