2024/G2/2

2.2 THE LIMIT OF A FUNCTION

Notes go here for 2.2... example:
$\lim _{z\to z}[f(x)g(x])=c\cdot [\lim _{x\to a}f(x)]$

2.3 CALCULATING LIMITS USING THE LIMIT LAWS

$\lim _{x\to a}=c\cdot [\lim _{x\to a}f(x)]$

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

Horizontal Asymptote or H.A
$f(x)=\lim _{x\to \infty }{\frac {x^{2}-1}{x^{2}+1}}=\lim _{x\to \infty }{\frac {x^{2}-1}{x^{2}+1}}\cdot {\frac {\frac {1}{x^{2}}}{\frac {1}{x^{2}}}}=\lim _{x\to \infty }{\frac {1-{\frac {1}{x^{2}}}}{1+{\frac {1}{x^{2}}}}}={\frac {1-0}{1+0}}=1$
This ultimately means that the equation will APPROACH TO ONE it will not ever be on one.
$\lim _{x\to \infty }{\frac {1}{x}}=0\lim _{x\to -\infty }{\frac {1}{x}}=0$
$\lim _{x\to \infty }{\frac {3x^{2}-x-2}{5x^{2}+4x+1}}=\lim _{x\to \infty }({\frac {3x^{2}-x-2}{5x^{2}+4x+1}})\cdot {\frac {\frac {1}{x^{2}}}{\frac {1}{x^{2}}}}=\lim _{x\to \infty }{\frac {3-{\frac {1}{x}}-{\frac {2}{x^{2}}}}{5+{\frac {4}{x}}+{\frac {1}{x^{2}}}}}={\frac {3-(0)-(0)}{5+(0)+(0)}}={\frac {3}{5}}$
V.A or a vertical asymptote is when the function becomes infinity such as $\lim _{x\to c}f(x)=\infty$

2.7 DERIVATIVES AND RATES OF CHANGE

To find the Tangent Line we use $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$
We later apply the points on which we want to find the slope. $f(x)=x^{2}f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {(x+h)^{2}-x^{2}}{h}}=\lim _{h\to 0}{\frac {x^{2}+2xh+h^{2}-x^{2})}{h}}=\lim _{h\to 0}{\frac {h^{2}+2xh}{h}}=\lim _{h\to 0}{\frac {({\cancel {h}})(h+2x)}{\cancel {h}}}$
$=\lim _{h\to 0}h+2x=(0)+2x$

2.8 THE DERIVATIVE AS A FUNCTION

${\frac {d}{dx}}[c]=0$
${\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]$

${\frac {d}{dx}}[f(x)\pm g(x)]={\frac {d}{dx}}[f(x)]\pm {\frac {d}{dx}}[g(x)]$
${\frac {d}{dx}}[x^{n}]=n\cdot x^{(n-1)}$
${\frac {d}{dx}}[a^{x}]=\ln(a)a^{x}$
${\frac {d}{dx}}[e^{x}]=e^{x}$