# Show that lim_(x to +oo)f'(x)=0 ?

## Given $f$ differentiable in $\mathbb{R}$ with $f \left(x\right) \ne 0$ $\forall x$$\in$$\mathbb{R}$. If ${\lim}_{x \to + \infty} f \left(x\right)$ exists then show that ${\lim}_{x \to + \infty} f ' \left(x\right) = 0$

Jan 25, 2018

See below.

#### Explanation:

Solved it.

${\lim}_{x \to + \infty} f \left(x\right)$$\in$$\mathbb{R}$

Supposed lim_(xto+oo)f(x)=λ

then ${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{{e}^{x} f \left(x\right)}{e} ^ x$

We have $\left(\frac{\pm \infty}{+ \infty}\right)$ and $f$ is differentiable in $\mathbb{R}$ so applying Rules De L'Hospital:

${\lim}_{x \to + \infty} \frac{{e}^{x} f \left(x\right)}{e} ^ x =$

${\lim}_{x \to + \infty} \frac{{e}^{x} f \left(x\right) + {e}^{x} f ' \left(x\right)}{e} ^ x =$

${\lim}_{x \to + \infty} \left(\frac{{e}^{x} f \left(x\right)}{e} ^ x + \frac{{e}^{x} f ' \left(x\right)}{e} ^ x\right) =$

${\lim}_{x \to + \infty} \left[f \left(x\right) + f ' \left(x\right)\right]$ =λ

• $h \left(x\right) = f \left(x\right) + f ' \left(x\right)$ with lim_(xto+oo)h(x)=λ

Thus, $f ' \left(x\right) = h \left(x\right) - f \left(x\right)$

Therefore, ${\lim}_{x \to + \infty} f ' \left(x\right) = {\lim}_{x \to + \infty} \left[h \left(x\right) - f \left(x\right)\right]$

=λ-λ=0

As a result,

${\lim}_{x \to + \infty} f ' \left(x\right) = 0$