Limits - find parameters γ, δ

**theintervurt** · February 5th 2013, 08:49 AM

Made a new thread for this as instructed by Plato. On to the limit itself

$\lim _{x\mapsto 0} \frac {\gamma + \delta \sin {x}}{x^2} = 3$

I found this in a book with exercises and although Plato mentioned it has no solutions the book says they're γ = 6 and δ = -6 . Also I double checked it and there is no x in front of γ .

**Plato** · February 5th 2013, 09:29 AM

Originally Posted by theintervurt

Made a new thread for this as instructed by Plato. On to the limit itself

$\lim _{x\mapsto 0} \frac {\gamma + \delta \sin {x}}{x^2} = 3$

I found this in a book with exercises and although Plato mentioned it has no solutions the book says they're γ = 6 and δ = -6 . Also I double checked it and there is no x in front of γ .

That practice book is completely wrong.
There is no solution to that limit as written.

It must be a typo in that text.

**theintervurt** · February 5th 2013, 10:04 AM

Forgive me I just realized it's not a sine but a cosine up there, I learn everything by reading greek books and the english equivalents and terminology is new to me so I might get some of it wrong.

**Soroban** · February 5th 2013, 12:08 PM

Hello, theintervurt!

We need these two facts:

. . $1 - \cos\theta \:=\:2\sin^2\tfrac{\theta}{2}$

. . $\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

$\lim _{x\to 0} \frac {a + b\cos x}{x^2} \:=\: 3$

$\text{We have: }\:f(x) \:=\:\frac {a + b\cos x}{x^2}$

$\text{Factor out }a\!:\;\;f(x) \:=\: \frac{a\left(1 + \frac{b}{a}\cos x\right)}{x^2}{$

$\text{Let }\tfrac{b}{a} = \text{-}1\!:\;f(x) \:=\:\frac{a(1 - \cos x)}{x^2}$

$\text{Then: }\:f(x) \:=\:\frac{a\cdot 2\sin^2\frac{x}{2}}{x^2} \:=\:\frac{\frac{1}{4}\left(2a\sin^2\frac{x}{2} \right)}{\frac{1}{4}(x^2)} \:=\:\tfrac{1}{2}{a}\frac{\sin^2\frac{x}{2}}{( \frac{x}{2})^2} \:=\:\frac{a}{2}\left(\frac{\sin\frac{x}{2}}{ \frac{x}{2}}\right)^2$

$\text{Since }\lim_{x\to0} f(x) \:=\:3,\,\text{ we have:}$

. . $\lim_{x\to0}\frac{a}{2}\left(\frac{\sin\frac{x}{2} }{\frac{x}{2}}\right)^2 \:=\:3 \quad\Rightarrow\quad \frac{a}{2}(1^2) \:=\:3 \quad\Rightarrow\quad \frac{a}{2} \:=\:3$

$\text{Therefore: }\:a = 6,\;b = \text{-}6$

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