# Limit calculation without Lhopital's rule

• Dec 14th 2011, 04:25 AM
noteiler1
Limit 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, 09:00 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

$\chi$ $\sigma$
• Dec 14th 2011, 09:58 AM
anonimnystefy
Re: Limit calculation without Lhopital's rule
hi chisigma

it's not Christmas here in Serbia!

and what about the second limit?
• Dec 14th 2011, 10:08 AM
noteiler1
Re: Limit calculation without Lhopital's rule
Oh, and i forgot to tell you the second condition, without using Taylor expansion.
• Dec 14th 2011, 10:43 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?
• Dec 14th 2011, 10:52 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 ?
• Dec 14th 2011, 01:00 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 $e^x - e$
• Dec 14th 2011, 04:23 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*}
• Dec 14th 2011, 04:33 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?
• Dec 14th 2011, 04:33 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*}
• Dec 14th 2011, 11:38 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,

$\lim_{x\to 1}\frac{e^x-e}{\sin (x^2-1)}=\lim_{x\to 1}\frac{e(x-1)}{ x^2-1}=\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}=\lim_{x\to 1}\frac{e}{x+1}=\frac{e}{2}$
• Dec 15th 2011, 01:17 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?
• Dec 15th 2011, 02:48 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.
• Dec 15th 2011, 03:06 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...
• Dec 15th 2011, 03:22 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

$\chi$ $\sigma$