Respuesta :
Answer:
Value of the limit is 0.5.
Step-by-step explanation:
Given,
[tex]F(x)=\frac{e^x-1-x}{x^2}[/tex]
When,
[tex]x=1,F(1)=frac{e^1-1-1}{1}=e-2=0.718281[/tex]
[tex]x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885[/tex]
[tex]x=0.1, F(0.1)=\frac{e^0.1-1-0.1}{(0.1)^2}=0.517091[/tex]
[tex]x=0.05, F(0.05)=\frac{e^0.05-1-0.05}{(0.05)^2}=0.508438[/tex]
[tex]x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)[/tex]
Correct upto six decimal places.
Now,
[tex]\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}[/tex] [tex](\frac{0}{0})[/tex] form, applying L-Hospital rule that is differentiating numerator and denominator we get,
[tex]\lim_{x\to 0}F(x)[/tex]
[tex]=\lim_{x\to 0}\frac{e^x-1}{2x}[/tex] [tex](\frac{0}{0})[/tex] form.
[tex]=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)[/tex]
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).
