Question

Evaluate the function at the given numbers( correct to six decimal places). Use the results to guess the value of the limit or explain why it does not exist. F(x)=e^x-1-x/x^2. X=positive and negative 1, 0.5, 0.1, 0.05, and 0.01. As the limit approaches 0.

Answer 1

Answer:

Value of the limit is 0.5.

Step-by-step explanation:

Given,

$F(x)=\frac{e^x-1-x}{x^2}$

When,

$x=1,F(1)=frac{e^1-1-1}{1}=e-2=0.718281$

$x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885$

$x=0.1, F(0.1)=\frac{e^0.1-1-0.1}{(0.1)^2}=0.517091$

$x=0.05, F(0.05)=\frac{e^0.05-1-0.05}{(0.05)^2}=0.508438$

$x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)$

Correct upto six decimal places.

Now,

$\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}$   $(\frac{0}{0})$ form, applying L-Hospital rule that is differentiating numerator and denominator we get,

$\lim_{x\to 0}F(x)$

$=\lim_{x\to 0}\frac{e^x-1}{2x}$    $(\frac{0}{0})$ form.

$=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)$

Limit exist and is 0.5. That is according to (1) we can see as the value of x lesser than 1 and tending to near 0, value of the function decreases respectively. And from (2) it shows ultimately it decreases and reach at 0.5, consider as limit point of F(x).