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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 $4x^2-6-4x^2\color{red}{+x}$?
Sep
28
comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$
I used the two things you showed me and progressed to $$\frac{-6-x}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4+\frac{1}{x}}}$$. While it does seem a bit better it is still not quite clear to me what to do with that.
Sep
28
comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$
Does the first identity have a name I can search for?