Limits with trigonometric functions

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Hello

$$\text{Is given:}\;\;\;\;\;\lim_{x\to0}\frac{\sqrt{\text{cos}x}-1}{x^2}$

$$A)\;0$

$$B)\;1$

$$C)\;\frac34$

$$D)\;-\frac14$

Attempted:

Actually I haven't done much, just not to say nothing at all, could be simple, but since I hate trig...

$$\lim_{x\to0}\frac{\sqrt{\text{cos}x}-1}{x^2}\cdot\frac{\sqrt{\text{cos}x}+1}{\sqrt{\text{cos}x}+1}=\lim_{x\to0}\frac{\text{cos}x-1}{x^2(\sqrt{\text{cos}x}+1)}=\text{Stuck!}$

royhaas · **Posted:** Fri, 24 Feb 2012 18:17:45 UTC

A series expansion will work, or you can use 1-cos(x) = 2sin^2(x/2)

.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

royhaas wrote:

A series expansion will work, or you can use 1-cos(x) = 2sin^2(x/2)

.

Ufff...All I can do is to change numerator into:

$$\lim_{x\to0}\frac{\text{cos}(x)-1}{x^2(\sqrt{\text{cos}(x)}+1)}\;\;\;\;\longrightarrow\;\;\;\lim_{x\to0}\frac{-1\left[2\cdot\text{sin}^2\left(\frac{x}{2}\right)\right]}{x^2(\sqrt{\text{cos}(x)}+1)}$

I feel dumb. Still don't know what to do next :oops:

any further step?

Soroban · **Posted:** Fri, 24 Feb 2012 22:37:30 UTC

Hello, kreshnik!

Quote:

$$\lim_{x\to0}\frac{\sqrt{\text{cos}x}-1}{x^2}$

. . $(A)\;0 \qquad (B)\;1 \qquad (C)\;\frac34 \qquad (D)\;-\frac14$

Multiply by $\frac{\sqrt{\cos x} + 1}{\sqrt{\cos x} + 1}:\;\;\dfrac{\sqrt{\cos x} - 1}{x^2} \cdot\dfrac{\sqrt{\cos x}+1}{\sqrt{\cos x} + 1} \;=\;\dfrac{\cos x - 1}{x^2(\sqrt{\cos x} + 1)} \;=\; \dfrac{-(1-\cos x)}{x^2(1+\sqrt{\cos x})}$

Multiply by $\frac{1+\cos x}{1+\cos x}:\;\;\dfrac{-(1-\cos x)}{x^2(1+\sqrt{\cos x})} \cdot\dfrac{1+\cos x}{1+\cos x} \;=\;-\dfrac{1-\cos^2x}{x^2(1 + \sqrt{\cos x})(1 + \cos x)}$

. . . . . . . . . . . . . . $=\;-\dfrac{\sin^2x}{x^2(1 + \sqrt{\cos x})(1 + \cos x)} \;=\;-\left(\dfrac{\sin x}{x}\right)^2\cdot\dfrac{1}{(1 + \sqrt{\cos x})(1 + \cos x)}$

$$\text{Therefore: }\:\lim_{x\to0}\left[ -\left(\dfrac{\sin x}{x}\right)^2\cdot\dfrac{1}{(1+\sqrt{\cos x})(1 + \cos x)}\right] \;=\; -(1^2)\left(\frac{1}{1+1}\right)\left(\frac{1}{1+1}\right) \;=\;-\frac{1}{4}$

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Soroban wrote:

Hello, kreshnik!

Quote:

$$\lim_{x\to0}\frac{\sqrt{\text{cos}x}-1}{x^2}$

. . $(A)\;0 \qquad (B)\;1 \qquad (C)\;\frac34 \qquad (D)\;-\frac14$

Multiply by $\frac{\sqrt{\cos x} + 1}{\sqrt{\cos x} + 1}:\;\;\dfrac{\sqrt{\cos x} - 1}{x^2} \cdot\dfrac{\sqrt{\cos x}+1}{\sqrt{\cos x} + 1} \;=\;\dfrac{\cos x - 1}{x^2(\sqrt{\cos x} + 1)} \;=\; \dfrac{-(1-\cos x)}{x^2(1+\sqrt{\cos x})}$

Multiply by $\frac{1+\cos x}{1+\cos x}:\;\;\dfrac{-(1-\cos x)}{x^2(1+\sqrt{\cos x})} \cdot\dfrac{1+\cos x}{1+\cos x} \;=\;-\dfrac{1-\cos^2x}{x^2(1 + \sqrt{\cos x})(1 + \cos x)}$

. . . . . . . . . . . . . . $=\;-\dfrac{\sin^2x}{x^2(1 + \sqrt{\cos x})(1 + \cos x)} \;=\;-\left(\dfrac{\sin x}{x}\right)^2\cdot\dfrac{1}{(1 + \sqrt{\cos x})(1 + \cos x)}$

$$\text{Therefore: }\:\lim_{x\to0}\left[ -\left(\dfrac{\sin x}{x}\right)^2\cdot\dfrac{1}{(1+\sqrt{\cos x})(1 + \cos x)}\right] \;=\; -(1^2)\left(\frac{1}{1+1}\right)\left(\frac{1}{1+1}\right) \;=\;-\frac{1}{4}$

Many thanks, I'm speechless. Thank you

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

An alternate approach using first order approximations. cos(x) ≈ 1 - (x^2)/2.
From this √cos(x) ≈ 1 - (x^2)/4 and numerator ≈ -(x^2)/4, so limit = -1/4.

outermeasure · **Posted:** Sat, 25 Feb 2012 03:07:55 UTC

kreshnik wrote:

royhaas wrote:

A series expansion will work, or you can use 1-cos(x) = 2sin^2(x/2)

.

Ufff...All I can do is to change numerator into:

$$\lim_{x\to0}\frac{\text{cos}(x)-1}{x^2(\sqrt{\text{cos}(x)}+1)}\;\;\;\;\longrightarrow\;\;\;\lim_{x\to0}\frac{-1\left[2\cdot\text{sin}^2\left(\frac{x}{2}\right)\right]}{x^2(\sqrt{\text{cos}(x)}+1)}$

I feel dumb. Still don't know what to do next :oops:

any further step?

Split the fractions up. No need to multiply by $\dfrac{1+\cos x}{1+\cos x}$ :
$\displaystyle \underbrace{\frac{-2}{1+\sqrt{\cos x}}}_{\to\dots}\cdot\underbrace{\left[\frac{\sin\left(\frac{x}{2}\right)}{x}\right]^2}_{\to\dots}\to\dots$

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 992

outermeasure wrote:

Split the fractions up. No need to multiply by $\dfrac{1+\cos x}{1+\cos x}$ :
$\displaystyle \underbrace{\frac{-2}{1+\sqrt{\cos x}}}_{\to\dots}\cdot\underbrace{\left[\frac{\sin\left(\frac{x}{2}\right)}{x}\right]^2}_{\to\dots}\to\dots$

I agree...no need to encumber yourself with more algebra than necessary.
Remember to change variables in the second limit: $$t=\frac{x}{2}$ , etc.

S.O.S. Mathematics CyberBoard

Limits with trigonometric functions

Who is online