Question

we want to find the limit
$f(x) = \frac{e^{4.2x} - e^{2.5x}}{x}$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{4.2x} - e^{2.5x}}{x}$.
start by calculating the values of the function for the inputs listed in this table.

$x$	$f(x)$
$0.1$
$0.05$
$0.01$
$0.001$
$0.0001$
$0.00001$

based on the values in this table, it appears
$\lim_{x \to 0} \frac{e^{4.2x} - e^{2.5x}}{x} = $

Explanation:

Step1: Substitute x=0.2 into f(x)

$f(0.2)=\frac{e^{4.2\times0.2}-e^{2.5\times0.2}}{0.2}=\frac{e^{0.84}-e^{0.5}}{0.2}\approx\frac{2.3164 - 1.6487}{0.2}\approx3.3385$

Step2: Substitute x=0.1 into f(x)

$f(0.1)=\frac{e^{4.2\times0.1}-e^{2.5\times0.1}}{0.1}=\frac{e^{0.42}-e^{0.25}}{0.1}\approx\frac{1.5220 - 1.2840}{0.1}\approx2.3800$

Step3: Substitute x=0.05 into f(x)

$f(0.05)=\frac{e^{4.2\times0.05}-e^{2.5\times0.05}}{0.05}=\frac{e^{0.21}-e^{0.125}}{0.05}\approx\frac{1.2337 - 1.1331}{0.05}\approx2.0120$

Step4: Substitute x=0.01 into f(x)

$f(0.01)=\frac{e^{4.2\times0.01}-e^{2.5\times0.01}}{0.01}=\frac{e^{0.042}-e^{0.025}}{0.01}\approx\frac{1.0429 - 1.0253}{0.01}\approx1.7600$

Step5: Substitute x=0.001 into f(x)

$f(0.001)=\frac{e^{4.2\times0.001}-e^{2.5\times0.001}}{0.001}=\frac{e^{0.0042}-e^{0.0025}}{0.001}\approx\frac{1.00421 - 1.00250}{0.001}\approx1.7100$

Step6: Substitute x=0.0001 into f(x)

$f(0.0001)=\frac{e^{4.2\times0.0001}-e^{2.5\times0.0001}}{0.0001}=\frac{e^{0.00042}-e^{0.00025}}{0.0001}\approx\frac{1.0004201 - 1.0002500}{0.0001}\approx1.7010$

Step7: Substitute x=0.00001 into f(x)

$f(0.00001)=\frac{e^{4.2\times0.00001}-e^{2.5\times0.00001}}{0.00001}=\frac{e^{0.000042}-e^{0.000025}}{0.00001}\approx\frac{1.0000420009 - 1.0000250003}{0.00001}\approx1.7001$

Step8: Identify the limit trend

As $x$ approaches 0, $f(x)$ approaches 1.7.

Answer:

$x$	$f(x)$
0.1	2.3800
0.05	2.0120
0.01	1.7600
0.001	1.7100
0.0001	1.7010
0.00001	1.7001

$\lim_{x
ightarrow0} \frac{e^{4.2x}-e^{2.5x}}{x}=1.7$