What is lim_(x->0) ln(sin x + cos x)/x ?

1 Answer
Mar 9, 2017

${\lim}_{x \to 0} \ln \frac{\sin x + \cos x}{x} = 1$

Explanation:

$\textcolor{w h i t e}{}$
Intuitive analysis

Notice that for small values of $x$:

$\sin x \approx x$

$\cos x \approx 1$

$\ln \left(x + 1\right) \approx x$

So:

$\ln \left(\sin x + \cos x\right) \approx \ln \left(x + 1\right) \approx x$

So we should expect:

${\lim}_{x \to 0} \ln \frac{\sin x + \cos x}{x} = 1$

$\textcolor{w h i t e}{}$
Semi-formal analysis

sin x = sum_(n=0)^oo ((-1)^n)/((2n+1)!) x^(2n+1) = x+O(x^3)

cos x = sum_(n=0)^oo ((-1)^n)/((2n)!) x^(2n) = 1+O(x^2)

$\ln \left(x + 1\right) = {\sum}_{n = 0}^{\infty} \frac{{\left(- 1\right)}^{n}}{n + 1} {x}^{n + 1} = x + O \left({x}^{2}\right)$

So:

$\ln \frac{\sin x + \cos x}{x} = \ln \frac{x + 1 + O \left({x}^{2}\right)}{x}$

$\textcolor{w h i t e}{\ln \frac{\sin x + \cos x}{x}} = \frac{x + O \left({x}^{2}\right)}{x}$

$\textcolor{w h i t e}{\ln \frac{\sin x + \cos x}{x}} = 1 + O \left(x\right) \to 1 \text{ }$ as $x \to 0$