MHB Limit involving a hyperbolic function

[Yankel]

Hello all,

I am trying to solve a limit:

\[\lim_{x\rightarrow 0}\frac{sinh (x)}{x}\]

I found many suggestions online, from complex numbers to Taylor approximations.

Finally I found a reasonable solution, but one move there doesn't make sense to me.

I am attaching a picture:

View attachment 9401

I have marked in blue the move I can't understand. How does the minus from the exponent goes down before the half ?

And one more thing, at the final move, I know that the limit is 1, I know it as a rule, but why is this rule apply ? Thank you !

 

[topsquark]

Hello all,

I am trying to solve a limit:

\[\lim_{x\rightarrow 0}\frac{sinh (x)}{x}\]

I found many suggestions online, from complex numbers to Taylor approximations.

Finally I found a reasonable solution, but one move there doesn't make sense to me.

I am attaching a picture:
I have marked in blue the move I can't understand. How does the minus from the exponent goes down before the half ?

And one more thing, at the final move, I know that the limit is 1, I know it as a rule, but why is this rule apply ? Thank you !

They are employing a trick that only works in this one limit. (ie. You can't do this for anything but [math]x \to 0[/math]. ) Since everything is nice and continuous:
[math]\lim_{x \to 0} e^{-x} = \lim_{x \to 0} e^x[/math]

-Dan

 

[I like Serena]

I have marked in blue the move I can't understand. How does the minus from the exponent goes down before the half ?

Hi Yankel,

It's a substitution. It's a bit confusing that they substitute with the same letter though. I recommend not doing that.
Anyway, substitute $x=-u$ and we get:
$$-\frac 12 \lim_{x\to 0} \frac{e^{-x}-1}{x} = -\frac 12 \lim_{u\to 0} \frac{e^{u}-1}{-u} = +\frac 12 \lim_{u\to 0} \frac{e^{u}-1}{u}
$$
Now replace $u$ by $x$ to get:
$$+\frac 12 \lim_{u\to 0} \frac{e^{u}-1}{u}=+\frac 12 \lim_{x\to 0} \frac{e^{x}-1}{x}$$

And one more thing, at the final move, I know that the limit is 1, I know it as a rule, but why is this rule apply ? Thank you !

It's L'Hôpital's rule. Both the numerator and the denominator approach 0. In such indeterminate cases we can replace both by their derivative to get:
$$\lim_{x\to 0} \frac{e^{x}-1}{x} = \lim_{x\to 0} \frac{(e^{x}-1)'}{x'} = \lim_{x\to 0} \frac{e^{x}}{1} = 1$$
Alternatively we can recognize the definition of a derivative and see:
$$\frac d{dx}(e^x)\big|_{x=0}=\lim_{h\to 0} \frac{e^{0+h}-e^0}{h-0}=\lim_{x\to 0} \frac{e^{x}-1}{x} = e^0=1$$

For the record, we could also have applied L'Hôpital's rule immediately at the beginning and get:
$$\lim_{x\to 0}\frac{\sinh(x)}{x} = \lim_{x\to 0}\frac{\sinh'(x)}{x'}= \lim_{x\to 0}\frac{\cosh(x)}{1}=\frac{\cosh(0)}1=1$$

 

[skeeter]

recall the definition of a derivative at a specific value of $x=a$ in the domain of $f(x)$ ...

$\displaystyle f'(a) = \lim_{x \to a} \dfrac{f(x) - f(a)}{x-a}$

for $f(x) = e^x$ ...

$\displaystyle \lim_{x \to 0} \dfrac{e^x - 1}{x} = \lim_{x \to 0} \dfrac{e^x - e^0}{x-0} = \lim_{x \to 0} \dfrac{f(x) - f(0)}{x-0} = f'(0) = 1$

for $g(x) = e^{-x}$ ...

$\displaystyle \lim_{x \to 0} \dfrac{e^{-x} - 1}{x} = \lim_{x \to 0} \dfrac{e^{-x} - e^0}{x-0} = \lim_{x \to 0} \dfrac{g(x) - g(0)}{x-0} = g'(0) = -1$

Therefore ...

$\displaystyle \dfrac{1}{2}\bigg[ \lim_{x \to 0} \dfrac{e^x - 1}{x} - \lim_{x \to 0} \dfrac{e^{-x} - 1}{x} \bigg] = \dfrac{1}{2}[1 - (-1)] = 1$

 

[Prove It]

Hello all,

I am trying to solve a limit:

\[\lim_{x\rightarrow 0}\frac{sinh (x)}{x}\]

I found many suggestions online, from complex numbers to Taylor approximations.

Finally I found a reasonable solution, but one move there doesn't make sense to me.

I am attaching a picture:
I have marked in blue the move I can't understand. How does the minus from the exponent goes down before the half ?

And one more thing, at the final move, I know that the limit is 1, I know it as a rule, but why is this rule apply ? Thank you !

The most concise way in this case, since it is a $\displaystyle \frac{0}{0}$ indeterminate form, is to use L'Hospital's Rule.

$\displaystyle \begin{align*} \lim_{x \to 0} \frac{\sinh{ \left( x \right) }}{x} &= \lim_{x \to 0} \frac{\cosh{\left( x \right) }}{1} \\ &= \lim_{x \to 0} \cosh{ \left( x \right) } \\ &= \cosh{ \left( 0 \right) } \\ &= 1 \end{align*}$