(without using l'hopital's rule) how to do you prove that:
lim x->0 (sinx)/x=1?
lim x->0 (tan3x)/x =3?
lim x->0 (1-cosx)x?
(without using l'hopital's rule) how to do you prove that:
lim x->0 (sinx)/x=1 ?
$$\small{\text{
$
\boxed{ \lim\limits_{x\to0}
\left(
\dfrac{ \sin{(x)} } { x } \right)
\qquad \sin (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!} = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\dotsb
} % boxed
$
}}$\\\\\\$
\small{\text{
$
\lim\limits_{x\to0}\left(
\dfrac{ \sin{(x)} } { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
\dfrac{x}{1!}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}\mp\dotsb
} { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
x \left( \dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right)
} { x } \right)
$
}}$\\\\\\$
\small{\text{
$
=
\lim\limits_{x\to0}\left(
\dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right)= \dfrac{1}{1!} = \textcolor[rgb]{1,0,0}{1}
$
}}$$
(without using l'hopital's rule) how to do you prove that:
lim x->0 (tan3x)/x =3 ?
$$\small{\text{
$
\boxed{ \lim\limits_{x\to0}
\left(
\dfrac{ \tan{(3x)} } { x } \right)
\qquad \tan{ (3x) } &= (3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb
} % boxed
$
}}$\\\\\\$
\small{\text{
$
\lim\limits_{x\to0}\left(
\dfrac{ \tan{(3x)} } { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
(3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb
} { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
x \left( 3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right)
} { x } \right)
$
}}$\\\\\\$
\small{\text{
$
=
\lim\limits_{x\to0}\left(
3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right)= \textcolor[rgb]{1,0,0}{3}
$
}}$$