Approximate $\frac{a^h-1}{h}$ah−1h for different values of $a$a by filling in the gaps in the table below. Round your answers to six decimal places where necessary.
$h$h | $2$2 | $e$e | $5$5 |
---|---|---|---|
$1$1 | $1$1 | $1.718282$1.718282 | $\editable{}$ |
$0.1$0.1 | $\editable{}$ | $\editable{}$ | $\editable{}$ |
$0.01$0.01 | $0.695555$0.695555 | $1.005017$1.005017 | $1.622459$1.622459 |
$0.001$0.001 | $0.693387$0.693387 | $1.000500$1.000500 | $1.610734$1.610734 |
$0.0001$0.0001 | $0.693171$0.693171 | $1.000050$1.000050 | $1.609567$1.609567 |
$0.00001$0.00001 | $\editable{}$ | $\editable{}$ | $\editable{}$ |
For which of these values of $a$a is $\lim_{h\to0}\left(\frac{a^h-1}{h}\right)=1$limh→0(ah−1h)=1?
$2$2
$e$e
$5$5
None of these three values.
We wish to determine the gradient function of $f\left(x\right)=e^x$f(x)=ex by first principles.
The value of $e^x$ex, where $e$e is the natural base, can be given by the expansion below:
$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\text{. . .}$ex=1+x1!+x22!+x33!+x44!+x55!+. . .
Consider the function $f\left(x\right)=e^x$f(x)=ex.