Hi everybody, i have problems with solving these limits without using Lhopital's rule. Problems are in attachment.
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)
Marry Christmas from Serbia
$\displaystyle \chi$ $\displaystyle \sigma$
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 $\displaystyle e^x - e$
$\displaystyle \displaystyle \begin{align*} \lim_{x \to 1}{\frac{e^x - e}{\sin{\left(x^2 - 1\right)}}} &= \lim_{x \to 1}\frac{e^x}{2x\cos{\left(x^2 - 1\right)}} \textrm{ by L'Hospital's Rule} \\ &= \frac{e^1}{2\cdot 1 \cdot \cos{\left(1^2 - 1\right)}} \\ &= \frac{e}{2} \end{align*}$
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,
$\displaystyle \lim_{x\to 1}\frac{e^x-e}{\sin (x^2-1)}=\lim_{x\to 1}\frac{e(x-1)}{ x^2-1}=\lim_{x\to 1}\frac{e(x-1)}{(x+1)(x-1)}=\lim_{x\to 1}\frac{e}{x+1}=\frac{e}{2}$
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?
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)
Marry Christmas from Serbia
$\displaystyle \chi$ $\displaystyle \sigma$