Hi everybody, i have problems with solving these limits without using Lhopital's rule. Problems are in attachment.

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- Dec 14th 2011, 03:25 AMnoteiler1Limit calculation without Lhopital's rule
Hi everybody, i have problems with solving these limits without using Lhopital's rule. Problems are in attachment.

- Dec 14th 2011, 08:00 AMchisigmaRe: Limit calculation without Lhopital's rule
Let's start from the limit...

$\displaystyle \lim_{x \rightarrow 1} \frac{e^{x}-e}{\sin (x^{2}-1}$ (1)

First we construct the Taylor expansions...

$\displaystyle e^{x}-e= e\ (x-1)\ \{1 + \frac{x-1}{2}+...\}$ (2)

$\displaystyle \sin (x^{2}-1) = (x-1)\ \{ (x+1) - \frac{(x-1)^{2}\ (x+1)^{3}}{3!} +...\}$ (3)

... and then we devide (2) and (3) obtaining the result...

$\displaystyle \lim_{x \rightarrow 1} \frac{e^{x}-e}{\sin (x^{2}-1}= \frace{e}{2}$ (4)

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$ - Dec 14th 2011, 08:58 AManonimnystefyRe: Limit calculation without Lhopital's rule
hi chisigma

it's not Christmas here in Serbia!

and what about the second limit? - Dec 14th 2011, 09:08 AMnoteiler1Re: Limit calculation without Lhopital's rule
Oh, and i forgot to tell you the second condition, without using Taylor expansion.

- Dec 14th 2011, 09:43 AMSironRe: Limit calculation without Lhopital's rule
Why do you want to solve them without using l'Hopital's rule or taylor expansion?

- Dec 14th 2011, 09:52 AMAmerRe: Limit calculation without Lhopital's rule
- Dec 14th 2011, 12:00 PMnoteiler1Re: Limit calculation without Lhopital's rule
I need to solve them without Lhopital's rule and Taylor expansion, becouse one of my friends has these problems during first semester of studying maths, and they don't know these technics yet. I solved couple of similar problems using the fact,

that if you have $\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}$ then you can substitute $\displaystyle f(x)$ with $\displaystyle h(x)$ when

$\displaystyle \lim_{x \to a} h(x) = 0$ and

$\displaystyle \lim_{x \to a} \frac{f(x)}{h(x)}=1$ and then calculate

$\displaystyle \lim_{x \to a} \frac{h(x)}{g(x)}$. I can find these functions for

$\displaystyle \sin(\frac{x^2}{\pi})$ and $\displaystyle \sin(x^2 -1)$ but this doesn't help if i don't know these functions for $\displaystyle 2^{\sqrt{\sin(x)+1}}-2$

and $\displaystyle e^x - e$ - Dec 14th 2011, 03:23 PMProve ItRe: Limit calculation without Lhopital's rule
$\displaystyle \displaystyle \begin{align*} \lim_{x \to 1}{\frac{e^x - e}{\sin{\left(x^2 - 1\right)}}} &= \lim_{x \to 1}\frac{e^x}{2x\cos{\left(x^2 - 1\right)}} \textrm{ by L'Hospital's Rule} \\ &= \frac{e^1}{2\cdot 1 \cdot \cos{\left(1^2 - 1\right)}} \\ &= \frac{e}{2} \end{align*}$

- Dec 14th 2011, 03:33 PMPlatoRe: Limit calculation without Lhopital's rule
- Dec 14th 2011, 03:33 PMProve ItRe: Limit calculation without Lhopital's rule
- Dec 14th 2011, 10:38 PMFernandoRevillaRe: Limit calculation without Lhopital's rule
Use the linear approximation of $\displaystyle f(x)$ by means of its tangent line. For example, for $\displaystyle f(x)=e^x-e$ the equation of the tangent line at $\displaystyle (1,0)$ is $\displaystyle y=e(x-1)$ . So,

$\displaystyle \lim_{x\to 1}\frac{e^x-e}{\sin (x^2-1)}=\lim_{x\to 1}\frac{e(x-1)}{ x^2-1}=\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}=\lim_{x\to 1}\frac{e}{x+1}=\frac{e}{2}$ - Dec 15th 2011, 12:17 AMnoteiler1Re: Limit calculation without Lhopital's rule
Yes, indeed $\displaystyle \lim_{x\to 1}\frac{e^x-e}{e(x-1)}=1$ and

$\displaystyle \lim_{x \to 1} \frac{\sin(x^2-1)}{x^2-1}=1$ so we can substitute and calculate $\displaystyle \lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}$. How couldn't i see it!! Thanks alot. But what about second problem? - Dec 15th 2011, 01:48 AMFernandoRevillaRe: Limit calculation without Lhopital's rule
- Dec 15th 2011, 02:06 AMProve ItRe: Limit calculation without Lhopital's rule
- Dec 15th 2011, 02:22 AMchisigmaRe: Limit calculation without Lhopital's rule
An alternative approach [I do hope it is 'allowed'...] to the second integral uses the 'infinite product'...

$\displaystyle \sin z = z\ \prod_{n=1}^{\infty} (1-\frac{z^{2}}{\pi^{2}\ n^{2}})$ (1)

Setting in (1) $\displaystyle z=x^{2}-1$ and with a little computation we obtain...

$\displaystyle \frac{e^{x}-e}{\sin (x^{2}-1)}= e\ \frac{e^{x-1}-1}{x-1}}\ \frac{1}{(x+1)\ \prod_{n=1}^{\infty} (1-\frac{(x^{2}-1)^{2}}{\pi^{2}\ n^{2}})}$ (2)

... and from (2), taking into account the 'fundamental limit'...

$\displaystyle \lim_{x \rightarrow 0} \frac{e^{x}-1}{x}=1$ (3)

... we obtain [finally!]...

$\displaystyle \lim_{x \rightarrow 1} \frac{e^{x}-e}{\sin (x^{2}-1)}= \frac{e}{2}$ (4)

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$