2
$\begingroup$

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}

$\endgroup$
  • 1
    $\begingroup$ As an addition you may want to consider this answer as well, in particular the second part of it $\endgroup$ – roman Dec 17 '18 at 11:38
2
$\begingroup$

From here we have

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ah... brainfart. Thanks. I'll accept as answer once the time limit is up. $\endgroup$ – Quaz Dec 17 '18 at 7:44
  • $\begingroup$ @Quaz You are welcome bye! $\endgroup$ – user Dec 17 '18 at 7:46
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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

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

$$=?$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.