limits?

**sbsite** · January 4th 2007, 01:30 PM

Help with some questions?

**ThePerfectHacker** · January 4th 2007, 03:21 PM

Problem #2

Expand $(2x-3)^{20}$ as a binomial,
$(2^{20}x^{20}+....)$

Expand $(3x+2)^{30}$ as a binomial,
$(3^{30}x^{30}+....)$

Expand $(2x+1)^{50}$ as a binomial,
$(2^{50}x^{50}+....)$

Thus, we have,
$\lim_{x\to \infty} \frac{(2^{20}x^{20}+...)(3^{30}x^{30}+...)}{2^{50} x^{50}+...}$
Take the Cauchy product,
$\lim_{x\to\infty} \frac{2^{20}\cdot 3^{30}x^{50}+...}{2^{50}x^{50}+...}$
Only the first coefficients are of interest,
$\frac{2^{20}\cdot 3^{30}}{2^{50}}$
Thus,
$(3/2)^{50}$

**galactus** · January 4th 2007, 04:05 PM

For #3, we can use the ol' Hospital rule.

$\lim_{x\rightarrow{0}}\frac{(8+3x-x^{2})^{\frac{1}{3}}-2}{x+x^{2}}$

L'Hopital, derivative of num. and den.:

$\frac{d}{dx}[(8+3x-x^{2})^{\frac{1}{3}}-2]=\frac{3-2x}{3(8+3x-x^{2})^{\frac{2}{3}}}$

$\frac{d}{dx}[x+x^{2}]=2x+1$

$\frac{\frac{3-2x}{3(8+3x-x^{2})^{\frac{2}{3}}}}{2x+1}=\frac{3-2x}{3(2x+1)(8+3x-x^{2})^{\frac{2}{3}}}$

$\frac{1}{3}\lim_{x\rightarrow{0}}\frac{3-2x}{(8+3x-x^{2})^{\frac{2}{3}}(1+2x)}$

= $\frac{1}{3}\cdot\frac{3}{8^{\frac{2}{3}}}=\frac{1} {4}$

For #6:

There are some identities we can employ:

Believe it or not: $\frac{sin(5x)-sin(3x)}{sin(x)}=2cos(4x)$

Now, we have a nice, little, itty-bitty, easy limit.

$\lim_{x\rightarrow{0}}2cos(4x)=2$

**topsquark** · January 5th 2007, 04:28 AM

For number 1 we can try L'Hopital's rule since the numerator and denominator both go to 0:
$\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt[3]{1+x} - \sqrt[3]{1-x}} = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}} - \frac{-1}{2\sqrt{1-x}}}{\frac{1}{3\sqrt[3]{(1+x)^2}} - \frac{-1}{3\sqrt[3]{(1-x)^2}}}$ $= \frac{\frac{1}{2} + \frac{1}{2}}{\frac{1}{3} + \frac{1}{3}} = \frac{1}{\frac{2}{3}}$ $= \frac{3}{2}$

-Dan

**topsquark** · January 5th 2007, 04:44 AM

For 5 there are other ways to do this (like L'Hopital's rule), but you can use one of my favorite methods. Since we are looking at x close to 0, use a Taylor's expansion of the numerator and denominator around x = 0:
$tgx - sin(x) \approx (tg - 1)x - \frac{1}{6}x^3 + \frac{1}{120}x^5$

$sin^3(x) \approx x^3 - \frac{1}{2}x^5$

So:
$\lim_{x \to 0} \frac{tgx - sin(x)}{sin^3(x)} = \lim_{x \to 0} \frac{(tg - 1)x - \frac{1}{6}x^3 + \frac{1}{120}x^5}{x^3 - \frac{1}{2}x^5}$

$= \lim_{x \to 0} \frac{x[(tg - 1) - \frac{1}{6}x^2 + \frac{1}{120}x^4]}{x[x^2 - \frac{1}{2}x^4]}$

$= \lim_{x \to 0} \frac{(tg - 1) - \frac{1}{6}x^2 + \frac{1}{120}x^4}{x^2 - \frac{1}{2}x^4}$

Now, the denominator goes to 0 while the numerator goes to $tg - 1$ . Presuming that $tg - 1 \neq 0$ the limit is infinite (ie doesn't exist.)

If $tg - 1 = 0$ then

$\lim_{x \to 0} \frac{tgx - sin(x)}{sin^3(x)} = \lim_{x \to 0} \frac{- \frac{1}{6}x^2 + \frac{1}{120}x^4}{x^2 - \frac{1}{2}x^4}$

$= \lim_{x \to 0} \frac{x^2[-\frac{1}{6} + \frac{1}{120}x^2]}{x^2[1 - \frac{1}{2}x^2]}$

$= \lim_{x \to 0} \frac{-\frac{1}{6} + \frac{1}{120}x^2}{1 - \frac{1}{2}x^2}$

$= \frac{-\frac{1}{6}}{1} = - \frac{1}{6}$

-Dan

**topsquark** · January 5th 2007, 04:48 AM

#4. I don't have time for full rigor here, but since $\sqrt{x + \sqrt{x + \sqrt{x}}} > \sqrt{x}$ for all x > 1 I would say this limit is infinite. To get a proof, note that the limit is of the form $\infty - \infty$ . You can convert this to either $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and then use L'Hopital's rule.

-Dan

**Soroban** · January 5th 2007, 05:22 AM

Hello, sbsite!

Here's #2 . . . without the Binomial Expansion.

$2)\;\lim_{x\to\infty}\frac{(2x-3)^{20}(3x+2)^{30}}{(2x+1)^{50}}$

Divide top and bottom by $(2x)^{50}$

. . $\frac{\frac{1}{(2x)^{50}}}{\frac{1}{(2x)^{50}}}\cd ot\frac{(2x-3)^{20}(3x+2)^{30}}{(2x+1)^{50}}\;=\;\frac{ \frac{(2x-3)^{20}}{(2x)^{20}}\,\frac{(3x+2)^{30}}{(2x)^{30}} } {\frac{(2x+1)^{50}}{(2x)^{50}}}$

. . $=\;\frac{\left(\frac{2x-3}{2x}\right)^{20}\left(\frac{3x+2}{2x}\right)^{30 } } {\left(\frac{2x+1}{2x}\right)^{50}} \;=\;\frac{\left(1 - \frac{3}{2x}\right)^{20}\left(\frac{3}{2} + \frac{1}{x}\right)^{30} } {\left(1 + \frac{1}{2x}\right)^{50}}$

Then: . $\lim_{x\to\infty}\frac{\left(1 - \frac{3}{2x}\right)^{20}\left(\frac{3}{2} + \frac{1}{x}\right)^{30} } {\left(1 + \frac{1}{2x}\right)^{50}}\;=\;\frac{(1 - 0)^{20}\left(\frac{3}{2} + 0\right)^{30}}{(1 + 0)^{50}}$ $=\;\frac{1^{20}\left(\frac{3}{2}\right)^{30}}{1^{5 0}} \:=\:\boxed{\left(\frac{3}{2}\right)^{30}}$

**Soroban** · January 5th 2007, 06:01 AM

Hello again, sbsite!

A couple more . . . without using L'Hopital.

$5)\;\lim_{x\to0}\frac{\tan x - \sin x}{\sin^3\!x}$

We have: . $\frac{\tan x - \sin x}{\sin^3\!x} \;=\;\frac{\frac{\sin x}{\cos x} - \sin x}{\sin^3\!x} \;=\;\frac{\frac{1}{\cos x} - 1}{\sin^2\!x}$ $=\;\frac{1 - \cos x}{\sin^2\!x\cos x}$

Multiply top and bottom by $1 + \cos x\!:\;\;\frac{1 - \cos x}{\sin^2\!x\cos x}\cdot\frac{1 + \cos x}{1 + \cos x}$

. . $=\;\frac{1 - \cos^2x}{\sin^2\!x\cos x(1 + \cos x)} \;= \;\frac{\sin^2\!x}{\sin^2\!x\cos x(1 + \cos x)} \;=\;\frac{1}{\cos x(1 + \cos x)}$

Therefore: . $\lim_{x\to0}\,\frac{1}{\cos x(1 + \cos x)} \;=\;\frac{1}{1(1+1)} \;=\;\boxed{\frac{1}{2}}$

$6)\;\lim_{x\to0}\frac{\sin 5x - \sin 3x}{\sin x}$

Divide top and bottom by $x\!:\;\;\frac{\frac{\sin 5x}{x} - \frac{\sin3x}{x}}{\frac{\sin x}{x}}$

Multiply the first fraction by $\frac{5}{5}$ , the second fraction by $\frac{3}{3}$

. . $\frac{\frac{5}{5}\!\cdot\!\frac{\sin 5x}{x} - \frac{3}{3}\!\cdot\!\frac{\sin 3x}{x}}{\frac{\sin x}{x}} \;=\;\frac{5\left(\frac{\sin 5x}{5x}\right) - 3\left(\frac{\sin3x}{3x}\right)}{\frac{\sin x}{x}}$

Then: . $\lim_{x\to0}\,\frac{5\left(\frac{\sin5x}{5x}\right ) - 3\left(\frac{\sin3x}{3x}\right)}{\frac{\sin x}{x}} \;=\;\frac{5\!\cdot\!1 - 3\!\cdot\!1}{1} \;= \;\boxed{2}$

**topsquark** · January 5th 2007, 06:33 AM

Originally Posted by topsquark

#4. I don't have time for full rigor here, but since $\sqrt{x + \sqrt{x + \sqrt{x}}} > \sqrt{x}$ for all x > 1 I would say this limit is infinite. To get a proof, note that the limit is of the form $\infty - \infty$ . You can convert this to either $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and then use L'Hopital's rule.

-Dan

#4 Addendum. I was thinking to factor a $\sqrt{x}$ from the first term and factor, and use the fact that the limit now takes the form of $\infty \cdot 0$ . You can convert this to the form $\frac{\infty}{\infty}$ and use L'Hopital's rule on it, but we now get an indeterminate form $\frac{0}{0}$ , and it doesn't look another application of L'Hopital's rule is going to help.

Perhaps this will work and I'm simply not being persistant enough, but I'm assuming another method (which I can't think of at the moment) will produce easier results.

(Note: Numerical estimates suggest that the limit is $\frac{1}{2}$ .)

-Dan

**TD!** · January 5th 2007, 06:52 AM

For #7:

Substitute x = y + 1, then you have to take the limit for y to 0.

Note that tan(pi/2+a) = -cot(a), so we can simplify like this:

$ \mathop {\lim }\limits_{x \to 1} \left( {1 - x} \right)\tan \left( {\frac{{\pi x}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} \left( { - y} \right)\tan \left( {\frac{{\pi \left( {1 + y} \right)}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} y\cot \left( {\frac{{\pi y}}{2}} \right) $

Now, arround 0, Taylor says that $\tan x \approx x$ ; so:

$ \mathop {\lim }\limits_{y \to 0} y\cot \left( {\frac{{\pi y}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} \frac{y}{{\tan \left( {\frac{{\pi y}}{2}} \right)}} = \mathop {\lim }\limits_{y \to 0} \frac{y}{{\frac{{\pi y}}{2}}} = \boxed{\frac{2}{\pi }} $

**TD!** · January 5th 2007, 07:16 AM

For #4:

To avoid l'Hopital with all these roots, multiply numerator and denominator with the complement:

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

Simplify in the numerator ((a-b)(a+b) = a²-b²) and divide numerator and denominator by sqrt(x):

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

Keep shifting the x in the square roots (sqrt(a)/x = sqrt(a/x²) with x positive):

$ \frac{{\sqrt {\sqrt {\frac{{x^2 + x}}{{x^2 }}} } }}{{\sqrt {1 + \sqrt {\frac{{x + \sqrt x }}{{x^2 }}} } + 1}} = \frac{{\sqrt {\sqrt {1 + \frac{1}{x}} } }}{{\sqrt {1 + \sqrt {\frac{1}{x} + \frac{1}{{x^{\frac{3}{2}} }}} } + 1}} $

Now, all the fractions 1/x and the one 1/x^(3/2) go to zero for x to infinity, so:

$ \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {\sqrt {1 + \frac{1}{x}} } }}{{\sqrt {1 + \sqrt {\frac{1}{x} + \frac{1}{{x^{\frac{3}{2}} }}} } + 1}} = \frac{{\sqrt {\sqrt {1 + 0} } }}{{\sqrt {1 + \sqrt {0 + 0} } + 1}} = \frac{1}{{1 + 1}} = \frac{1}{2} $

**TD!** · January 5th 2007, 10:47 AM

Addendum for #7.

If you can't (or rather not) use Taylor, the last steps can be replaced by using the standard limit of tan(x)/x for x to 0.
That limit is of course, just as the one for sin(x)/x with x to 0, 1. In that case, write:

$ \mathop {\lim }\limits_{y \to 0} \frac{y}{{\tan \left( {\frac{\pi }{2}y} \right)}} = \mathop {\lim }\limits_{y \to 0} \frac{2}{\pi }\frac{{\frac{\pi }{2}y}}{{\tan \left( {\frac{\pi }{2}y} \right)}} = \frac{2}{\pi }\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\tan \left( {\frac{\pi }{2}y} \right)}}{{\frac{\pi }{2}y}}} \right)^{ - 1} = \frac{2}{\pi } $

**galactus** · January 5th 2007, 02:45 PM

Here's a method I used for the nested radical. It's not as eloquent as TD's, but what the heck:

$\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}$

Let $u=\sqrt{x+\sqrt{x}}$

With some algbraic gymnastics, we get:

$x=\frac{(\sqrt{4u^{2}+1}-1)^{2}}{4}$

and $\frac{\sqrt{4u^{2}+1}-1}{2}$

And we have:

$\underbrace{\lim_{x\rightarrow{\infty}}\sqrt{\frac {1}{2}+u+u^{2}-\frac{\sqrt{4u^{2}+1}}{2}}}$ $-\underbrace{\lim_{x\rightarrow{\infty}}\frac{\sqrt {4u^{2}+1}}{2}}$ $+\lim_{x\rightarrow{\infty}}\boxed{\frac{1}{2}}$

............................................ $\underbrace{\nwarrow\nearrow}_{\text{tend to 0}}$

**Soroban** · January 5th 2007, 03:52 PM

Hello, sbsite!

I think I have #4 . . .

$4)\;\lim_{x\to+\infty}\left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}\right)$

Let $\sqrt{x} = u\quad\Rightarrow\quad x = u^2$

We have: . $\lim_{u\to\infty}\left(\sqrt{u^2 + \sqrt{u^2 + u}} - u\right)$

Multiply top and bottom by $\left(\sqrt{u^2 + \sqrt{u^2+u}} + u\right)$

. . $\frac{\sqrt{u^2 + \sqrt{u^2 + u}} - u}{1}\cdot\frac{\sqrt{u^2 + \sqrt{u^2 + u}} + u} {\sqrt{u^2 + \sqrt{u^2+u}} + u}$ $\:=\:\frac{u^2 + \sqrt{u^2 + u} - u^2}{\sqrt{u^2 + \sqrt{u^2 + u}} + u}$ $\:=\:\frac{\sqrt{u^2 + u}}{\sqrt{u^2 + \sqrt{u^2+u}} + u}$

Divide top and bottom by $u$

. . $\frac{\frac{\sqrt{u^2+u}}{u}} {\frac{\sqrt{u^2 + \sqrt{u^2+u}}}{u} + \frac{u}{u}}$ . $=\:\frac{\frac{\sqrt{u^2+u}}{\sqrt{u^2}}} {\frac{\sqrt{u^2 + \sqrt{u^2+u}}} {\sqrt{u^2}} + 1}$ . $= \:\frac{\sqrt{\frac{u^2+u}{u^2}}} {\sqrt{\frac{u^2 + \sqrt{u^2 + u}}{u^2}} + 1}$ . $= \:\frac{\sqrt{\frac{u^2}{u^2} + \frac{u}{u^2}}} {\sqrt{\frac{u^2}{u^2} + \frac{\sqrt{u^2 + u}}{u^2}} + 1}$

We have: . $\lim_{u\to\infty} \frac{\sqrt{1 + \frac{1}{u}}} {\sqrt{1 + \frac{\sqrt{u^2+u}}{u^2}} + 1}\;=\;\frac{\sqrt{1 + 0}}{\sqrt{1 + 0} + 1} \;=\;\boxed{\frac{1}{2}}$

**TD!** · January 6th 2007, 01:21 AM

Soroban, we did exactly the same, only you named sqrt(x) = u. The steps are the same

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