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Oct
1
comment Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital
How come $\frac{\sin x}{\sin 2x}\to \frac{1}{2}$? And also, I'm not sure I quite grasped the last line. I know that $\cos x \to 1$, but what does that do anyway?
Oct
1
comment Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital
It appears I am lacking on trigonometric properties. How did you go from $(\sin x - \tan x)$ to $\sin x(1 - \cos x)$?
Oct
1
asked Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital
Oct
1
comment What did I do wrong solving $\lim_{x\to0}\frac{2-\sqrt[6]{3x+64}}{5x}$?
@Lucian the factorization was missing a $160$ that's all.
Oct
1
accepted How do I solve $\lim_{x\to1}\frac{\sqrt{2-x}-1}{1+\sqrt[5]{x-2}}$?
Oct
1
comment How do I solve $\lim_{x\to1}\frac{\sqrt{2-x}-1}{1+\sqrt[5]{x-2}}$?
Ahh... My numerator factorization was wrong. It had to be $(w-1)(w^4+w^3+w^2+w+1)$, so at end it would indeed be $\frac{-5}{2}$. Thanks.
Oct
1
comment How do I solve $\lim_{x\to1}\frac{\sqrt{2-x}-1}{1+\sqrt[5]{x-2}}$?
Here's my attempt:$$w^{10} = 2-x \text{ therefore } w \to 1$$ then $$\lim_{w\to1}\frac{\sqrt{w^{10}} - 1}{1 + \sqrt[5]{-w^{10}}} = \frac{w^5-1}{1-w^2}$$ That is factorized to $$\frac{(w-1)(w^4+w^3+w^2+w)}{(w-1)(-w-1)}$$However $$\frac{1+1+1+1}{-1 - 1} = -2$$
Oct
1
asked How do I solve $\lim_{x\to1}\frac{\sqrt{2-x}-1}{1+\sqrt[5]{x-2}}$?
Oct
1
accepted What did I do wrong solving $\lim_{x\to0}\frac{2-\sqrt[6]{3x+64}}{5x}$?
Oct
1
comment What did I do wrong solving $\lim_{x\to0}\frac{2-\sqrt[6]{3x+64}}{5x}$?
This is it! Thank you..... it's the little things that mess everything up.
Oct
1
asked What did I do wrong solving $\lim_{x\to0}\frac{2-\sqrt[6]{3x+64}}{5x}$?
Oct
1
comment How do I rationalize higher index roots?
@BrevanEllefsen Sorry, I have edited my question now.
Oct
1
revised How do I rationalize higher index roots?
added 106 characters in body
Oct
1
asked How do I rationalize higher index roots?
Sep
30
comment How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$?
Usually when solving limits with infinity, I take the highest exponents. For example, with $$\lim_{x\to-\infty}\frac{2x+3}{x+\sqrt[3]{x}}$$ I would just take $$\frac{2x}{x}$$ Because they are the "highest exponents", resulting in $2$. For this exercise, given $$\sqrt{x+a} - \sqrt{x}$$, can I "take the highest exponents" like $$\sqrt{x} - \sqrt{x}$$ and conclude that $$\sqrt{x} - \sqrt{x} = 0$$? (Completely ignoring $a$ since it has no $x$ with exponent).
Sep
30
accepted Calculating $\lim_{x\to2^+}\frac{3x}{\ln(x-2)}$
Sep
30
comment How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$?
Would it have been acceptable to just say "if we grab the highest exponents we have $\sqrt{x} - \sqrt{x}$ = 0"? Given that $a$ is some real number, we could ignore it?
Sep
30
asked Calculating $\lim_{x\to2^+}\frac{3x}{\ln(x-2)}$
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.