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 Sep 30 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ I'm having trouble seeing how can $(-x-6) / |x|$ result in $(1 + 6/x)$. Since $|x| = +\infty$, which is positive, how come the signs are shifted? I mean, $-x$ becomes $1$ (negative to positive) and $-6$ becomes $6/x$ (negative to positive). Shouldn't $(-x-6)$ be divided by just $x$? Since $x = -\infty$ then the signs can indeed by shifted. I think. Sep 30 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Hello, today I worked on this limit and forgot to do the "inverting" thing. My result was $-\frac{1}{4}$ instead of $\frac{1}{4}$. I can see that I should have inverted that, but it is unclear to me why should I invert it. Whenever I see a $x\to-\infty$ do I have to change it to $x\to\infty$? Because without inverting it the operation went smoothly as far as I am aware of. Sep 30 comment Calculating $\lim_{x\to0^+}x-\frac{1}{x^3}$ @Travis ah yes, you are right. Sep 30 asked Calculating $\lim_{x\to0^+}x-\frac{1}{x^3}$ Sep 30 comment Is there a way I can test/verify my answers to limit computations? @RaziehNoori ah yes, but is there a way to verify this on paper? Without tools? Sep 30 asked Is there a way I can test/verify my answers to limit computations? Sep 30 comment Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Well that's a surprisingly clever way (to me) to approach this. Thanks! Sep 30 accepted Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Sep 30 comment Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ How did you get rid of the $x^2$ from $x^2\left(\frac{\sin x}{x}+1\right)$, and how did you divide $\left(\sin^2\frac{x}{2}\right)$ by $\frac{x}{2}$ near the end? Sep 30 comment Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ @ChocolateAndCheese for learning purposes I am supposed to be able to do this without L'Hopital. Sep 29 asked Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Sep 29 accepted How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$? Sep 29 comment How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$? Hmm...... So it's not really about replacing $x$ with $+\infty$, but rather consider what happens as $x$ keeps growing? I can see that as $x$ grows, the denominator will obviously grow further and the whole division will come closer to $0$. Sep 29 comment How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$? @SimonS well, I would have $$\frac{a}{\sqrt{x+a}+\sqrt{x}}$$ But wouldn't that cause the same problem? If $a = -\infty$. Sep 29 asked How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$? Sep 29 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ So basically a limit where $x\to-\infty$ is the same as $x\to+\infty$ as long as I invert all the $x$ signs? Of course $x^2$ would remain the same, but if I had $x^3$ it would become $-x^3$? Sep 29 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ When you divide $(-x-6)/|x|$, why did you pick $|x|$ instead of $x$? Why is this allowed? Also, how can $(-x-6)/|x|$ become $1+\frac{6}{x}$? Dividing by infinity should be $0$, I think. Sep 29 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ But isn't it $\sqrt{4x^2+x}$ instead of $\sqrt{4x^2-x}$? Sep 28 accepted Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Sep 28 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ After rationalizing, shouldn't it have been $4x^2-6-4x^2\color{red}{-x}$ instead of $4x^2-6-4x^2\color{red}{+x}$?