2

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

\begin{equation} \lim_{x\to\infty} \sqrt{x^2+1}-x+1 \end{equation}

\begin{equation} \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)} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{x^2+1 - x +1}{\sqrt{x^2+1}+x-1} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{x^2 - x +2}{\sqrt{x^2+1}+x-1} \end{equation}

\begin{equation} \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)} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{x - 1 +\frac{2}{x}}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{\infty - 1 + 0}{1+1-0} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{\infty - 1}{2} = \infty \end{equation}

Edit: Added correct steps for completeness, thanks for the quick answers!

\begin{equation} \lim_{x\to\infty} \sqrt{x^2+1}-x+1 \end{equation}

\begin{equation} \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)} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{x^2+1 - x^2+2x -1}{\sqrt{x^2+1}+x-1} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{2x}{\sqrt{x^2+1}+x-1} \end{equation}

\begin{equation} \lim_{x\to\infty} \frac{x}{x} \frac{2}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}} \end{equation}

\begin{equation} \frac{2}{1+1} = 1 \end{equation}

Quaz
  • 79

3 Answers3

3

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

Angina Seng
  • 158,341
2

From here we have

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

user
  • 154,566
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 $$ \lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}$$

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

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

$$=?$$

lab bhattacharjee
  • 274,582