Limit calculation without Lhopital's rule

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December 14th 2011, 03:25 AM
noteiler1

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Limit calculation without Lhopital's rule

Hi everybody, i have problems with solving these limits without using Lhopital's rule. Problems are in attachment.
December 14th 2011, 08:00 AM
chisigma

Re: Limit calculation without Lhopital's rule

Let's start from the limit...

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

First we construct the Taylor expansions...

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

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

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

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

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

Marry Christmas from Serbia

$\chi$ $\sigma$
December 14th 2011, 08:58 AM
anonimnystefy

Re: Limit calculation without Lhopital's rule

hi chisigma

it's not Christmas here in Serbia!

and what about the second limit?
December 14th 2011, 09:08 AM
noteiler1

Re: Limit calculation without Lhopital's rule

Oh, and i forgot to tell you the second condition, without using Taylor expansion.
December 14th 2011, 09:43 AM
Siron

Re: Limit calculation without Lhopital's rule

Why do you want to solve them without using l'Hopital's rule or taylor expansion?
December 14th 2011, 09:52 AM
Amer

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by noteiler1

Oh, and i forgot to tell you the second condition, without using Taylor expansion.

and without substitution ?
December 14th 2011, 12:00 PM
noteiler1

Re: 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 $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}$ then you can substitute $f(x)$ with $h(x)$ when
$\lim_{x \to a} h(x) = 0$ and
$\lim_{x \to a} \frac{f(x)}{h(x)}=1$ and then calculate
$\lim_{x \to a} \frac{h(x)}{g(x)}$ . I can find these functions for
$\sin(\frac{x^2}{\pi})$ and $\sin(x^2 -1)$ but this doesn't help if i don't know these functions for $2^{\sqrt{\sin(x)+1}}-2$
and $e^x - e$
December 14th 2011, 03:23 PM
Prove It

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by noteiler1

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

$\displaystyle \begin{align*} \lim_{x \to 1}{\frac{e^x - e}{\sin{\left(x^2 - 1\right)}}} &= \lim_{x \to 1}\frac{e^x}{2x\cos{\left(x^2 - 1\right)}} \textrm{ by L'Hospital's Rule} \\ &= \frac{e^1}{2\cdot 1 \cdot \cos{\left(1^2 - 1\right)}} \\ &= \frac{e}{2} \end{align*}$
December 14th 2011, 03:33 PM
Plato

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by Prove It

$\displaystyle \begin{align*} \lim_{x \to 1}{\frac{e^x - e}{\sin{\left(x^2 - 1\right)}}} &= \lim_{x \to 1}\frac{e^x}{2x\cos{\left(x^2 - 1\right)}} \textrm{ by L'Hospital's Rule} \\ &= \frac{e^1}{2\cdot 1 \cdot \cos{\left(1^2 - 1\right)}} \\ &= \frac{e}{2} \end{align*}$

@Prove It
Did you notice that the whole point of this thread is that the limits cannot be done with L'Hospital's Rule nor with series?
December 14th 2011, 03:33 PM
Prove It

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by noteiler1

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 $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}$ then you can substitute $f(x)$ with $h(x)$ when
$\lim_{x \to a} h(x) = 0$ and
$\lim_{x \to a} \frac{f(x)}{h(x)}=1$ and then calculate
$\lim_{x \to a} \frac{h(x)}{g(x)}$ . I can find these functions for
$\sin(\frac{x^2}{\pi})$ and $\sin(x^2 -1)$ but this doesn't help if i don't know these functions for $2^{\sqrt{\sin(x)+1}}-2$
and $e^x - e$

If you need help finding the derivative of $\displaystyle \begin{align*} 2^{\sqrt{\sin{x} + 1}} \end{align*}$ , remember that $\displaystyle \begin{align*} \frac{d}{dx}\left(a^x\right) &= a^x\ln{a} \end{align*}$
December 14th 2011, 10:38 PM
FernandoRevilla

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by noteiler1

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

Use the linear approximation of $f(x)$ by means of its tangent line. For example, for $f(x)=e^x-e$ the equation of the tangent line at $(1,0)$ is $y=e(x-1)$ . So,

$\lim_{x\to 1}\frac{e^x-e}{\sin (x^2-1)}=\lim_{x\to 1}\frac{e(x-1)}{ x^2-1}=\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}=\lim_{x\to 1}\frac{e}{x+1}=\frac{e}{2}$
December 15th 2011, 12:17 AM
noteiler1

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by FernandoRevilla

Use the linear approximation of $f(x)$ by means of its tangent line. For example, for $f(x)=e^x-e$ the equation of the tangent line at $(1,0)$ is $y=e(x-1)$ . So,

$\lim_{x\to 1}\frac{e^x-e}{\sin (x^2-1)}=\lim_{x\to 1}\frac{e(x-1)}{ x^2-1}=\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}=\lim_{x\to 1}\frac{e}{x+1}=\frac{e}{2}$

Yes, indeed $\lim_{x\to 1}\frac{e^x-e}{e(x-1)}=1$ and
$\lim_{x \to 1} \frac{\sin(x^2-1)}{x^2-1}=1$ so we can substitute and calculate $\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}$ . How couldn't i see it!! Thanks alot. But what about second problem?
December 15th 2011, 01:48 AM
FernandoRevilla

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by noteiler1

How couldn't i see it!! Thanks alot. But what about second problem?

Same way, find the corresponding tangent lines. Show your work (if you want), and we can check it.
December 15th 2011, 02:06 AM
Prove It

Re: Limit calculation without Lhopital's rule

Quote:

Originally Posted by Plato

@Prove It
Did you notice that the whole point of this thread is that the limits cannot be done with L'Hospital's Rule nor with series?

I must have misread the title...
December 15th 2011, 02:22 AM
chisigma

Re: Limit calculation without Lhopital's rule

An alternative approach [I do hope it is 'allowed'...] to the second integral uses the 'infinite product'...

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

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

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

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

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

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

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

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

Marry Christmas from Serbia

$\chi$ $\sigma$