Use the method of direct substitution. If, for example, we have lim x → 4 ( x + 4 ) {\displaystyle {\displaystyle \lim _{x\to 4}(x+4)}} {\displaystyle {\displaystyle \lim _{x\to 4}(x+4)}}, plug in 4 {\displaystyle 4} 4 where x {\displaystyle x} x is. That gives us 8 {\displaystyle 8} 8. The limit of f {\displaystyle f} f, where f ( x ) = x + 4 {\displaystyle f(x)=x+4} {\displaystyle f(x)=x+4}, at x = 4 {\displaystyle x=4} x=4 is 8 {\displaystyle 8} 8. This might not always work, though; when the problem involves rational functions with a variable in the denominator, like lim x → 2 x 2 − 4 x − 2 {\displaystyle \lim _{x\to 2}{\frac {\mathrm {x^{2}-4} }{\mathrm {x-2} }}} {\displaystyle \lim _{x\to 2}{\frac {\mathrm {x^{2}-4} }{\mathrm {x-2} }}}, substituting 2 {\displaystyle 2} {\displaystyle 2} for x {\displaystyle x} x will cause the function to equal 0 0 {\displaystyle {\frac {\mathrm {0} }{\mathrm {0} }}} {\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} {\displaystyle 0}, the limit does not exist.
Try factoring out and cancelling terms that lead to 0 {\displaystyle {\frac {\mathrm {0} }{\mathrm {0} }}} {\displaystyle {\frac {\mathrm {0} }{\mathrm {0} }}} or ∞ ∞ {\displaystyle {\frac {\mathrm {\infty } }{\mathrm {\infty } }}} {\displaystyle {\frac {\mathrm {\infty } }{\mathrm {\infty } }}}. In the previous example lim x → 2 x 2 − 4 x − 2 {\displaystyle \lim _{x\to 2}{\frac {\mathrm {x^{2}-4} }{\mathrm {x-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} {\displaystyle x-2}: lim x → 2 ( x − 2 ) ( x + 2 ) x − 2 {\displaystyle \lim _{x\to 2}{\frac {\mathrm {(x-2)(x+2)} }{\mathrm {x-2} }}} {\displaystyle \lim _{x\to 2}{\frac {\mathrm {(x-2)(x+2)} }{\mathrm {x-2} }}} = lim x → 2 ( x + 2 ) {\displaystyle {\displaystyle \lim _{x\to 2}(x+2)}} {\displaystyle {\displaystyle \lim _{x\to 2}(x+2)}}. We can evaluate it by plugging in 2 {\displaystyle 2} {\displaystyle 2} and the limit is 4 {\displaystyle 4} 4.
Try to multiply the numerator and the denominator with a conjugate. We have lim x → 4 2 − x 4 − x {\displaystyle \lim _{x\to 4}{\frac {\mathrm {2-{\sqrt {x}}} }{\mathrm {4-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}})} {\displaystyle (2+{\sqrt {x}})} will transform it into lim x → 4 4 − x ( 4 − x ) ( 2 + x ) {\displaystyle \lim _{x\to 4}{\frac {\mathrm {4-x} }{\mathrm {(4-x)(2+{\sqrt {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)} {\displaystyle (4-x)} to get a simpler lim x → 4 1 2 + x {\displaystyle \lim _{x\to 4}{\frac {\mathrm {1} }{\mathrm {2+{\sqrt {x}}} }}} {\displaystyle \lim _{x\to 4}{\frac {\mathrm {1} }{\mathrm {2+{\sqrt {x}}} }}}. This comes up to 1 4 {\displaystyle {\frac {\mathrm {1} }{\mathrm {4} }}} {\displaystyle {\frac {\mathrm {1} }{\mathrm {4} }}}.
Use trigonometric transformations. If your limit is lim θ → 0 1 − c o s θ θ {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {1-cos\theta } }{\mathrm {\theta } }}} {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {1-cos\theta } }{\mathrm {\theta } }}}, multiply the numerator and denominator with ( 1 + c o s θ ) {\displaystyle (1+cos\theta )} {\displaystyle (1+cos\theta )} to get lim θ → 0 1 − c o s 2 θ ( θ ) ( 1 + c o s θ ) {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {1-cos^{2}\theta } }{\mathrm {(\theta )(1+cos\theta )} }}} {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {1-cos^{2}\theta } }{\mathrm {(\theta )(1+cos\theta )} }}}. Use s i n 2 θ + c o s 2 θ = 1 {\displaystyle sin^{2}\theta +cos^{2}\theta =1} {\displaystyle sin^{2}\theta +cos^{2}\theta =1} and separate the multiplied fractions to obtain lim θ → 0 s i n θ {\displaystyle \lim _{\theta \to 0}sin\theta } {\displaystyle \lim _{\theta \to 0}sin\theta } ∗ {\displaystyle *} * s i n θ θ {\displaystyle {\frac {\mathrm {sin\theta } }{\mathrm {\theta } }}} {\displaystyle {\frac {\mathrm {sin\theta } }{\mathrm {\theta } }}} ∗ {\displaystyle *} * 1 ( 1 + c o s θ ) {\displaystyle {\frac {\mathrm {1} }{\mathrm {(1+cos\theta )} }}} {\displaystyle {\frac {\mathrm {1} }{\mathrm {(1+cos\theta )} }}}. You can plug in 0 {\displaystyle 0} {\displaystyle 0} to get 0 ∗ 1 ∗ 1 {\displaystyle 0*1*1} {\displaystyle 0*1*1}. The limit is 0 {\displaystyle 0} {\displaystyle 0}.
Find limits at infinity. lim x → 0 1 x {\displaystyle \lim _{x\to 0}{\frac {\mathrm {1} }{\mathrm {x} }}} {\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 = 1 x {\displaystyle y={\frac {\mathrm {1} }{\mathrm {x} }}} {\displaystyle y={\frac {\mathrm {1} }{\mathrm {x} }}}, you will see that y → ∞ {\displaystyle {\displaystyle y\to \infty }} {\displaystyle {\displaystyle y\to \infty }} as x → 0 {\displaystyle {\displaystyle x\to 0}} {\displaystyle {\displaystyle x\to 0}}.
Use L'Hôpital's rule. This is used for indeterminate forms like 0 0 {\displaystyle {\frac {\mathrm {0} }{\mathrm {0} }}} {\displaystyle {\frac {\mathrm {0} }{\mathrm {0} }}} or ∞ ∞ {\displaystyle {\frac {\mathrm {\infty } }{\mathrm {\infty } }}} {\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 lim x → c f ( x ) {\displaystyle {\displaystyle \lim _{x\to c}f(x)}} {\displaystyle {\displaystyle \lim _{x\to c}f(x)}} = lim x → c h ( x ) = 0 {\displaystyle {\displaystyle \lim _{x\to c}h(x)}=0} {\displaystyle {\displaystyle \lim _{x\to c}h(x)}=0} or lim x → c f ( x ) {\displaystyle {\displaystyle \lim _{x\to c}f(x)}} {\displaystyle {\displaystyle \lim _{x\to c}f(x)}} = lim x → c h ( x ) = ± ∞ {\displaystyle {\displaystyle \lim _{x\to c}h(x)}=\pm \infty } {\displaystyle {\displaystyle \lim _{x\to c}h(x)}=\pm \infty } and h ′ ( x ) ≠ 0 {\displaystyle {\displaystyle h'(x)\neq 0}} {\displaystyle {\displaystyle h'(x)\neq 0}} for all x ≠ c {\displaystyle {\displaystyle x\neq c}} {\displaystyle {\displaystyle x\neq c}} in I {\displaystyle I} I and if lim x → c f ′ ( x ) h ′ ( x ) {\displaystyle \lim _{x\to c}{\frac {\mathrm {f'(x)} }{\mathrm {h'(x)} }}} {\displaystyle \lim _{x\to c}{\frac {\mathrm {f'(x)} }{\mathrm {h'(x)} }}} exists, lim x → c f ( x ) h ( x ) = lim x → c f ′ ( 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)} }}} {\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, lim θ → 0 s i n θ θ {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {sin\theta } }{\mathrm {\theta } }}} {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {sin\theta } }{\mathrm {\theta } }}} = lim θ → 0 c o s θ 1 {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {cos\theta } }{\mathrm {1} }}} {\displaystyle \lim _{\theta \to 0}{\frac {\mathrm {cos\theta } }{\mathrm {1} }}} = 1 1 {\displaystyle {\frac {\mathrm {1} }{\mathrm {1} }}} {\displaystyle {\frac {\mathrm {1} }{\mathrm {1} }}} = 1 {\displaystyle 1} 1.
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