Limits by brute force (make-up exercise)

Text

Compute the following limits using the brute-force method (a table of values):

(a) $\displaystyle\lim_{x\to3}\frac{x^2-9}{x-3}$

(b) $\displaystyle\lim_{x\to+\infty}\frac{\sqrt{x^2+1}}{x}$

(c) $\displaystyle\lim_{x\to0}\frac{\sin(5x)}{\sin(3x)}$

Solution

(a) Table of values:

$x$	$3.1$	$3.01$	$2.99$
$(x^2-9)/(x-3)$	$6.1$	$6.01$	$5.99$

The values converge to $6$: indeed, for $x\neq3$, $\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3\to6$.

(b) Table of values:

$x$	$10$	$100$	$1000$
$\sqrt{x^2+1}/x$	$1.005$	$1.00005$	$1.0000005$

The values approach $1$ from above: the limit is $1^+$.

(c) Table of values:

$x$	$0.1$	$0.01$	$0.001$
$\sin(5x)/\sin(3x)$	$1.622$	$1.666$	$1.6666$

The values settle at $\dfrac53$, consistent with $\dfrac{\sin(5x)}{\sin(3x)}=\dfrac{5x}{3x}\cdot\dfrac{\sin(5x)/(5x)}{\sin(3x)/(3x)}\to\dfrac53$.

$\boxed{6;\ 1;\ \tfrac53}$