Solving Algebraic Limits (Form 1)

When you’re given a complex rational function, you use the fourth and final algebraic limit-finding technique. The technique of plugging fails, because you end up with a 0 in one of the denominators.

$\displaystyle If\quad n\quad \epsilon \quad Q\quad then\quad \lim _{ x\rightarrow a }{ \frac { { x }^{ n }-{ a }^{ n } }{ x-a } } \\ =\lim _{ x\rightarrow { a }^{ + } }{ \frac { { x }^{ n }-{ a }^{ n } }{ x-a } } \qquad \left[ \because \lim _{ x\longrightarrow a }{ f\left( x \right) } exists\therefore \lim _{ x\longrightarrow a }{ f\left( x \right) } =\lim _{ x\rightarrow { a }^{ + } }{ f(x) } \right] \\ =\lim _{ h\longrightarrow 0 }{ \frac { \left( a+h \right) -{ a }^{ n } }{ a+h-a } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { { a }^{ n }\left\{ \left( 1+\frac { h }{ a } \right) -1 \right\} }{ h } } \\ ={ a }^{ n }\lim _{ h\longrightarrow 0 }{ \frac { \left\{ 1+n.\frac { h }{ a } +\frac { n\left( n-1 \right) }{ 2! } \frac { { h }^{ 2 } }{ { a }^{ 2 } } +....-1 \right\} }{ h } } \quad \left[ Using{ \left( 1+x \right) }^{ 2 }=1+nx+\frac { n\left( n-1 \right) }{ 2! } { x }^{ 2 }+....... \right] \\ ={ a }^{ n }\lim _{ h\longrightarrow 0 }{ \left( \frac { n }{ a } +\frac { n\left( n-1 \right) }{ 2! } \frac { h }{ { a }^{ 2 } } +.... \right) } \\ ={ a }^{ n }.\frac { n }{ a } =n{ a }^{ n-1 }\\ \therefore \lim _{ x\rightarrow a }{ \frac { { x }^{ n }-{ a }^{ n } }{ x-a } } =n{ a }^{ n-1 }$

For Example

$\displaystyle =\lim _{ x\longrightarrow \infty }{ \frac { { x }^{ 2 }+5 }{ { x }^{ 2 }+4x+3 } } \\ Dividing\quad { N }^{ r }\quad and\quad { D }^{ r }\quad by\quad { x }^{ 2 }\\ =\lim _{ x\longrightarrow \infty }{ \frac { 1+\frac { 5 }{ { x }^{ 2 } } }{ 1+\frac { 4 }{ x } +\frac { 3 }{ { x }^{ 2 } } } } \\ =\frac { 1+0 }{ 1+0+0 } \\ \left( because\quad \frac { K }{ x } \longrightarrow 0,when\quad x\longrightarrow \infty \quad where\quad K\quad is\quad any\quad constant \right) [\\ =1$