Question

start by calculating the values of the function for the inputs listed in this ta
based on the values in this table, it appears
\\(\lim\limits_{x\to 0} \dfrac{e^{4.2x} - e^{2.5x}}{x} = 1.7\\)

Explanation:

Step1: Define the given function

The function is $f(x) = \frac{e^{4.2x} - e^{2.5x}}{x}$

Step2: Calculate f(0.1)

Substitute $x=0.1$:
$f(0.1) = \frac{e^{4.2\times0.1} - e^{2.5\times0.1}}{0.1} = \frac{e^{0.42} - e^{0.25}}{0.1}$
Calculate exponentials: $e^{0.42}\approx1.5220$, $e^{0.25}\approx1.2840$
$f(0.1)=\frac{1.5220 - 1.2840}{0.1}=\frac{0.238}{0.1}=2.38$

Step3: Calculate f(0.001)

Substitute $x=0.001$:
$f(0.001) = \frac{e^{4.2\times0.001} - e^{2.5\times0.001}}{0.001} = \frac{e^{0.0042} - e^{0.0025}}{0.001}$
Calculate exponentials: $e^{0.0042}\approx1.004209$, $e^{0.0025}\approx1.002503$
$f(0.001)=\frac{1.004209 - 1.002503}{0.001}=\frac{0.001706}{0.001}=1.706$ (rounded to 1.71 is acceptable, or more precisely 1.706)

Step4: Calculate f(0.0001)

Substitute $x=0.0001$:
$f(0.0001) = \frac{e^{4.2\times0.0001} - e^{2.5\times0.0001}}{0.0001} = \frac{e^{0.00042} - e^{0.00025}}{0.0001}$
Calculate exponentials: $e^{0.00042}\approx1.00042009$, $e^{0.00025}\approx1.00025003$
$f(0.0001)=\frac{1.00042009 - 1.00025003}{0.0001}=\frac{0.00017006}{0.0001}=1.7006$

Answer:

• For $x=0.1$: $2.38$
• For $x=0.001$: $1.71$ (or $1.706$)
• For $x=0.0001$: $1.7006$
• The limit $\lim_{x\to0}\frac{e^{4.2x}-e^{2.5x}}{x}=1.7$