Question

guess the value of the limit (if it exists) by evaluating the function at the given numbers. (it is suggested that you report answers accurate to at least six decimal places.) let (f(x)=\frac{cos(3x)-cos(11x)}{x^{2}}). we want to find the limit (lim_{x
ightarrow0}\frac{cos(3x)-cos(11x)}{x^{2}}). start by calculating the values of the function for the inputs listed in this table. based on the values in this table, it appears (lim_{x
ightarrow0}\frac{cos(3x)-cos(11x)}{x^{2}}) question help: message instructor submit question jump to answer

Explanation:

Step1: Evaluate for $x = 0.2$

Substitute $x=0.2$ into $f(x)=\frac{\cos(3x)-\cos(11x)}{x^{2}}$.
$\cos(3\times0.2)=\cos(0.6)\approx0.825336$, $\cos(11\times0.2)=\cos(2.2)\approx - 0.666367$.
$f(0.2)=\frac{0.825336-(-0.666367)}{(0.2)^{2}}=\frac{1.491703}{0.04}=37.292575$.

Step2: Evaluate for $x = 0.1$

Substitute $x = 0.1$ into $f(x)$.
$\cos(3\times0.1)=\cos(0.3)\approx0.955337$, $\cos(11\times0.1)=\cos(1.1)\approx0.453596$.
$f(0.1)=\frac{0.955337 - 0.453596}{(0.1)^{2}}=\frac{0.501741}{0.01}=50.174100$.

Step3: Evaluate for $x = 0.05$

Substitute $x=0.05$ into $f(x)$.
$\cos(3\times0.05)=\cos(0.15)\approx0.988795$, $\cos(11\times0.05)=\cos(0.55)\approx0.858990$.
$f(0.05)=\frac{0.988795 - 0.858990}{(0.05)^{2}}=\frac{0.129805}{0.0025}=51.922000$.

Step4: Evaluate for $x = 0.01$

Substitute $x = 0.01$ into $f(x)$.
$\cos(3\times0.01)=\cos(0.03)\approx0.999550$, $\cos(11\times0.01)=\cos(0.11)\approx0.994008$.
$f(0.01)=\frac{0.999550 - 0.994008}{(0.01)^{2}}=\frac{0.005542}{0.0001}=55.420000$.

Step5: Evaluate for $x = 0.001$

Substitute $x=0.001$ into $f(x)$.
$\cos(3\times0.001)=\cos(0.003)\approx0.999996$, $\cos(11\times0.001)=\cos(0.011)\approx0.999940$.
$f(0.001)=\frac{0.999996 - 0.999940}{(0.001)^{2}}=\frac{0.000056}{0.000001}=56.000000$.

Step6: Evaluate for $x = 0.0001$

Substitute $x = 0.0001$ into $f(x)$.
$\cos(3\times0.0001)=\cos(0.0003)\approx0.999999955$, $\cos(11\times0.0001)=\cos(0.0011)\approx0.99999939$.
$f(0.0001)=\frac{0.999999955 - 0.99999939}{(0.0001)^{2}}=\frac{0.000000565}{1\times10^{- 8}}=56.500000$.

Step7: Evaluate for $x = 0.00001$

Substitute $x=0.00001$ into $f(x)$.
$\cos(3\times0.00001)=\cos(0.00003)\approx0.99999999955$, $\cos(11\times0.00001)=\cos(0.00011)\approx0.9999999939$.
$f(0.00001)=\frac{0.99999999955 - 0.9999999939}{(0.00001)^{2}}=\frac{0.00000000565}{1\times10^{-10}}=56.500000$.

Based on the trend of the values in the table, as $x$ approaches $0$, the values of $f(x)$ seem to be approaching $56.5$.

Answer:

When $x = 0.2$, $f(x)=37.292575$; when $x = 0.1$, $f(x)=50.174100$; when $x = 0.05$, $f(x)=51.922000$; when $x = 0.01$, $f(x)=55.420000$; when $x = 0.001$, $f(x)=56.000000$; when $x = 0.0001$, $f(x)=56.500000$; when $x = 0.00001$, $f(x)=56.500000$; $\lim_{x
ightarrow0}\frac{\cos(3x)-\cos(11x)}{x^{2}}\approx56.5$