# Calculate lim_(xto1)(e^f(x)-1)/(x-1) ?

## Given $f : \mathbb{R} \to \mathbb{R}$ continuous at ${x}_{0} = 1$ with ${\lim}_{x \to 1} f \frac{x}{x - 1} = 4$

Jun 10, 2018

Jun 10, 2018

Any good?

#### Explanation:

${\lim}_{x \to 1} \frac{{e}^{f} \left(x\right) - 1}{x - 1}$

= lim_(xto1)((1 + f(x) + (f^2(x))/(2!) + ...)-1)/(x-1)

= lim_(xto1)( f(x) + (f^2(x))/(2!) + ...)/(x-1)

= lim_(xto1)( f(x))/(x-1) + lim_(xto1) ((f^2(x))/(2!) )/(x-1) + ...

= lim_(xto1)( f(x))/(x-1) + lim_(xto1) (f(x))/(2!) f(x)/(x-1) + ...

• ${\lim}_{x \to 1} f \frac{x}{x - 1} = 4 \implies {\lim}_{x \to 1} f \left(x\right) = 0$

$= 4 + 0 \cdot 4 + \ldots = 4$

Jun 10, 2018

For $x$ near $1$ , consider

$\left\{\begin{matrix}g \left(x\right) = f \frac{x}{x - 1} \text{ " \\ lim_(xto1)g(x)=4" }\end{matrix}\right.$ $\iff$

$\left\{\begin{matrix}f \left(x\right) = \left(x - 1\right) g \left(x\right) \text{ " \\ lim_(xto1)g(x)=4" }\end{matrix}\right.$

Hence, $f \left(1\right) = {\lim}_{x \to 1} f \left(x\right) = {\lim}_{x \to 1} \left(x - 1\right) g \left(x\right) = 4 \cdot 0 = 0$

$I = {\lim}_{x \to 1} \frac{{e}^{f} \left(x\right) - 1}{x - 1} = {\lim}_{x \to 1} \frac{{e}^{\left(x - 1\right) g \left(x\right)} - 1}{x - 1}$

${\lim}_{x \to 1} g \left(x\right) = 4$ so there is $a > 0$ for which we have $0 <$$a <$$g \left(x\right)$ when $x \to 1$

$=$ ${\lim}_{x \to 1} \frac{{e}^{\left(x - 1\right) g \left(x\right)} - 1}{\left(x - 1\right) \cdot g \left(x\right)} \cdot g \left(x\right)$

• ${\lim}_{x \to 1} \frac{{e}^{\left(x - 1\right) g \left(x\right)} - 1}{\left(x - 1\right) \cdot g \left(x\right)} =$

Substitute

$\left(x - 1\right) g \left(x\right) = t$
$x \to 1$
$t \to 0$

${\lim}_{t \to 0} \frac{{e}^{t} - 1}{t} {=}_{D L H}^{\left(\frac{0}{0}\right)} {\lim}_{t \to 0} \frac{\left({e}^{t} - 1\right) '}{\left(t\right) '}$

$= {\lim}_{t \to 0} {e}^{t} = {e}^{0} = 1$

Therefore $I = {\lim}_{x \to 1} \frac{{e}^{f} \left(x\right) - 1}{x - 1} = 1 \cdot 4 = 4$