Намирането на границите на функциите е основно понятие в смятането. Границите се използват за изследване на поведението на функция около определена точка. Изчислителните граници включват много методи и тази статия очертава някои от тях.
Стъпки
Стъпка 1. Използвайте метода на директно заместване
Ако например имаме limx → 4 (x+4) { displaystyle { displaystyle / lim _ {x / до 4} (x+4)}}
plug in 4{displaystyle 4}
where x{displaystyle x}
is. that gives us 8{displaystyle 8}
. the limit of f{displaystyle f}
where f(x)=x+4{displaystyle f(x)=x+4}
at x=4{displaystyle x=4}
is 8{displaystyle 8}
. this might not always work, though; when the problem involves rational functions with a variable in the denominator, like limx→2x2−4x−2{displaystyle \lim _{x\to 2}{frac {mathrm {x^{2}-4} }{mathrm {x-2} }}}
substituting 2{displaystyle 2}
for x{displaystyle x}
will cause the function to equal 00{displaystyle {frac {mathrm {0} }{mathrm {0} }}}
giving you an indeterminate form. or, if you get an undefined result where the numerator is a non-zero value and the denominator is 0{displaystyle 0}
the limit does not exist.
step 2. try factoring out and cancelling terms that lead to 0{displaystyle {frac {mathrm {0} }{mathrm {0} }}}
or ∞∞{displaystyle {frac {mathrm {infty } }{mathrm {infty } }}}
. in the previous example limx→2x2−4x−2{displaystyle \lim _{x\to 2}{frac {mathrm {x^{2}-4} }{mathrm {x-2} }}}
we can factor out and cancel x−2{displaystyle x-2}
: limx→2(x−2)(x+2)x−2{displaystyle \lim _{x\to 2}{frac {mathrm {(x-2)(x+2)} }{mathrm {x-2} }}}
= limx→2(x+2){displaystyle {displaystyle \lim _{x\to 2}(x+2)}}
. we can evaluate it by plugging in 2{displaystyle 2}
and the limit is 4{displaystyle 4}
step 3. try to multiply the numerator and the denominator with a conjugate
we have limx→42−x4−x{displaystyle \lim _{x\to 4}{frac {mathrm {2-{sqrt {x}}} }{mathrm {4-x} }}}
. if you multiply the numerator and denominator with (2+x){displaystyle (2+{sqrt {x}})}
will transform it into limx→44−x(4−x)(2+x){displaystyle \lim _{x\to 4}{frac {mathrm {4-x} }{mathrm {(4-x)(2+{sqrt {x}})} }}}
. you can cancel out (4−x){displaystyle (4-x)}
to get a simpler limx→412+x{displaystyle \lim _{x\to 4}{frac {mathrm {1} }{mathrm {2+{sqrt {x}}} }}}
. this comes up to 14{displaystyle {frac {mathrm {1} }{mathrm {4} }}}
step 4. use trigonometric transformations
if your limit is limθ→01−cosθθ{displaystyle \lim _{theta \to 0}{frac {mathrm {1-cos\theta } }{mathrm {theta } }}}
multiply the numerator and denominator with (1+cosθ){displaystyle (1+cos\theta)}
to get limθ→01−cos2θ(θ)(1+cosθ){displaystyle \lim _{theta \to 0}{frac {mathrm {1-cos^{2}\theta } }{mathrm {(theta)(1+cos\theta)} }}}
. use sin2θ+cos2θ=1{displaystyle sin^{2}\theta +cos^{2}\theta =1}
and separate the multiplied fractions to obtain limθ→0sinθ{displaystyle \lim _{theta \to 0}sin\theta }
∗{displaystyle *}
sinθθ{displaystyle {frac {mathrm {sin\theta } }{mathrm {theta } }}}
∗{displaystyle *}
1(1+cosθ){displaystyle {frac {mathrm {1} }{mathrm {(1+cos\theta)} }}}
. you can plug in 0{displaystyle 0}
to get 0∗1∗1{displaystyle 0*1*1}
. the limit is 0{displaystyle 0}
step 5. find limits at infinity
limx→01x{displaystyle \lim _{x\to 0}{frac {mathrm {1} }{mathrm {x} }}}
has a limit at infinity. it cannot be simplified to be a finite number. examine the graph of the function if this is the case. for the limit in the example, if you look at the graph of y=1x{displaystyle y={frac {mathrm {1} }{mathrm {x} }}}
you will see that y→∞{displaystyle {displaystyle y\to \infty }}
as x→0{displaystyle {displaystyle x\to 0}}
step 6. use l'hôpital's rule
this is used for indeterminate forms like 00{displaystyle {frac {mathrm {0} }{mathrm {0} }}}
or ∞∞{displaystyle {frac {mathrm {infty } }{mathrm {infty } }}}
. this rule states that for functions f and h differentiable on an open interval i except at a point c in i, if limx→cf(x){displaystyle {displaystyle \lim _{x\to c}f(x)}}
= limx→ch(x)=0{displaystyle {displaystyle \lim _{x\to c}h(x)}=0}
or limx→cf(x){displaystyle {displaystyle \lim _{x\to c}f(x)}}
= limx→ch(x)=±∞{displaystyle {displaystyle \lim _{x\to c}h(x)}=\pm \infty }
and h′(x)≠0{displaystyle {displaystyle h'(x)\neq 0}}
for all x≠c{displaystyle {displaystyle x\neq c}}
in i{displaystyle i}
and if limx→cf′(x)h′(x){displaystyle \lim _{x\to c}{frac {mathrm {f'(x)} }{mathrm {h'(x)} }}}
exists, limx→cf(x)h(x)=limx→cf′(x)h′(x){displaystyle \lim _{x\to c}{frac {mathrm {f(x)} }{mathrm {h(x)} }}=\lim _{x\to c}{frac {mathrm {f'(x)} }{mathrm {h'(x)} }}}
. this rule converts indeterminate forms to forms that can be easily evaluated. for example, limθ→0sinθθ{displaystyle \lim _{theta \to 0}{frac {mathrm {sin\theta } }{mathrm {theta } }}}
= limθ→0cosθ1{displaystyle \lim _{theta \to 0}{frac {mathrm {cos\theta } }{mathrm {1} }}}
= 11{displaystyle {frac {mathrm {1} }{mathrm {1} }}}
= 1{displaystyle 1}
