I found $\lim _{x\to \:0}\left(1-\frac{1}{x}\left(\sqrt{1+x^2}-1\right)\right)$ to be -infinity

Question

Here is what I tried to calculate:

$\lim _{x\to \:0}\left(1-\frac{1}{x}\left(\sqrt{1+x^2}-1\right)\right)$

When I calculated for 0-(I mean negative zero) I found it to be -infinity but the solution found it to be 1

Can anyone verify it for me please?

My solution:

I found the equation as it is to have an intermediate form(1 - 0/0) so I tried to remove that by simplifying it to

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

So I calculated limit to 0-(Negative zero) and I found it to be -infinity

Answer 1

Imho easiest way: $$\frac{1}{x}\left(\sqrt{1+x^2}-1\right)=\frac{x}{\sqrt{1+x^2}+1}\to0$$

Now about your solution: you have little mistake, it should be $$\frac{1}{x}\left(\sqrt{1+x^2}-1\right)=\text{sign(x)}\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}$$ so you obtain not $-\infty$ but $\infty-\infty$ which is undefined.

Answer 2

Your method is incorrect as it is of the $-\infty+\infty$ form and nothing can be said about it.Also $\frac{\sqrt{1+x^2}}{x}=\sqrt{1+\frac{1}{x^2}}$ need not be true (what happens when $x$ is negative?)

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let $x=\tan t$ we have to calculate $$\lim_{t\to 0}1-\frac{\sqrt{1+\tan^2t}-1}{\tan t}=\lim_{t \to 0}(1-\tan (t/2))=1$$

Answer 3

Using the standard limit properties: $$\begin{align} \lim_{x\to 0} (1 - \frac{1}{x}(\sqrt{x^2+1}-1)) &= \lim_{x\to 0}1 - \lim_{x\to 0} \frac{\sqrt{x^2+1} - 1}{x} \\ &= 1 - \lim_{x\to 0}\left( \frac{\sqrt{x^2+1} - 1}{x}\cdot \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}\right) \\ &= 1 - \lim_{x\to 0}\frac{x^2+1-1}{x(\sqrt{x^2+1}+1)} \\ &= 1 - \lim_{x\to 0}\frac{x}{\sqrt{x^2+1}+1} \\ &= 1 - \frac{\lim_{x\to 0}x}{\lim_{x\to 0}(\sqrt{x^2+1}+1)} \\ &= 1 - \frac{0}{\sqrt{0^2+1}+1} = 1 - 0 = 1 \end{align}$$

$\underline{\text{EDIT}}$: In response to a question in the comments, here's how to find the limit of the same function at infinity: $$\begin{align} \lim_{x\to \infty}\left( 1-\frac{1}{x}(\sqrt{x^2+1}-1)\right) & = \lim_{x\to\infty}1 - \lim_{x\to\infty}\left( \frac{\sqrt{x^2+1}}{x}-\frac{1}{x}\right) \\ &= 1 - \left( \lim_{x\to\infty}\frac{\sqrt{x^2+1}}{x} - \lim_{x\to\infty}\frac{1}{x} \right) \\ &= 1 - \left( \lim_{x\to\infty} \frac{\sqrt{x^2+1}}{x}-0 \right) \\ &= 1 - \lim_{x\to\infty} \frac{\sqrt{x^2\left( 1+\frac{1}{x^2}\right) }}{x} \\ &= 1 - \lim_{x\to\infty} \frac{x\sqrt{1+\frac{1}{x^2}}}{x} \\ &= 1 - \lim_{x\to\infty} \sqrt{1+\frac{1}{x^2}} = 1-1=0 \end{align}$$