Limits by brute force (a table of values)

Text

For each limit, build a table of values (at least three) and determine the limit:

(a) $\displaystyle\lim_{x\to0^{+}}\frac{\sin(3x)}{x}$

(b) $\displaystyle\lim_{x\to+\infty}\frac{2x^2-x}{x^2+3}$

(c) $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}$

Solution

(a) Table of values:

$x$	$0.1$	$0.01$	$0.001$
$\sin(3x)/x$	$2.955$	$2.9996$	$3.000$

The values approach $3$: using the standard limit, $\dfrac{\sin(3x)}{x}=3\cdot\dfrac{\sin(3x)}{3x}\to 3\cdot1=3$.

(b) Table of values:

$x$	$10$	$100$	$1000$
$(2x^2-x)/(x^2+3)$	$1.845$	$1.985$	$1.998$

The values approach $2$: as $x\to+\infty$ the ratio of the leading-degree terms dominates, $\dfrac{2x^2}{x^2}=2$.

(c) Table of values:

$x$	$0.1$	$0.01$	$0.001$
$(1-\cos x)/x^2$	$0.4996$	$0.5000$	$0.5000$

The values settle at $\dfrac12$, consistent with the standard limit $\dfrac{1-\cos x}{x^2}\to\dfrac12$.

$\boxed{3;\ 2;\ \tfrac12}$