Limits: (lesson 3 of 5)
Limit of Irrational Functions
Example 1:
Multiplying by a unity factor
lim x → a = x − b − a − b x 2 − a 2 , a > b \mathop {\lim }\limits_{x \to a} = \frac{\sqrt {x - b} - \sqrt {a - b} }{x^2 - a^2}, \ a > b x → a lim = x 2 − a 2 x − b − a − b , a > b Solution:
Multiply by 1 in the form of the numerator with a "+" sign substituted for a "-" sign:
x − b − a − b x 2 − a 2 x − b + a − b x − b + a − b = ( x − b ) − ( a − b ) ( x 2 − a 2 ) ( x − b + a − b ) = x − a ( x − a ) ( x + a ) ( x − b + a − b ) = 1 ( x + a ) ( x − b + a − b )
\frac{\sqrt {x - b} - \sqrt {a - b} }{x^2 - a^2} \frac{\sqrt {x - b} + \sqrt {a - b} }{\sqrt {x - b} + \sqrt {a - b} }
= \frac{(x - b) - (a - b)}{(x^2 - a^2)(\sqrt{x - b} + \sqrt{a - b})} \\
= \frac{x - a}{(x - a) (x + a)(\sqrt{x - b} + \sqrt{a - b})} = \frac{1}{(x + a)(\sqrt{x - b} + \sqrt{a - b})}
x 2 − a 2 x − b − a − b x − b + a − b x − b + a − b = ( x 2 − a 2 ) ( x − b + a − b ) ( x − b ) − ( a − b ) = ( x − a ) ( x + a ) ( x − b + a − b ) x − a = ( x + a ) ( x − b + a − b ) 1 Therefore,
lim x → a = x − b − a − b x 2 − a 2 = lim x → a 1 ( x + a ) ( x − b + a − b ) = 1 ( a + a ) ( a − b + a − b ) = 1 4 a a − b
\mathop {\lim }\limits_{x \to a} = \frac{{\sqrt {x - b} - \sqrt {a - b} }}{{{x^2} - {a^2}}} = \mathop {\lim }\limits_{x \to a} \frac{1}{{(x + a)(\sqrt {x - b} + \sqrt {a - b} )}}
= \frac{1}{{(a + a)(\sqrt {a - b} + \sqrt {a - b} )}} = \frac{1}{{4a\sqrt {a - b} }}
x → a lim = x 2 − a 2 x − b − a − b = x → a lim ( x + a ) ( x − b + a − b ) 1 = ( a + a ) ( a − b + a − b ) 1 = 4 a a − b 1 Please note that in the above examples, once the limit has been taken,
the limit symbol is removed and the fixed point is substituted for x. Prior to
that step, the limit symbol is needed. When doing pure algebra, the limit symbol is left off to avoid cluttering the math.
Example 2:
lim x → 9 x − 3 x − 9 = lim x → 9 x − 3 x − 9 ⋅ x + 3 x + 3 = lim x → 9 x − 9 ( x − 9 ) ( x + 3 ) = lim x → 9 1 x + 3 = 1 9 + 3 = 1 6
\begin{aligned}
&\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} = \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} \cdot \frac{{\sqrt x + 3}}{{\sqrt x + 3}} = \mathop {\lim }\limits_{x \to 9} \frac{{x - 9}}{{(x - 9)(\sqrt x + 3)}} = \\
&\mathop {\lim }\limits_{x \to 9} \frac{1}{{\sqrt x + 3}} = \frac{1}{{\sqrt 9 + 3}} = \frac{1}{6}
\end{aligned}
x → 9 lim x − 9 x − 3 = x → 9 lim x − 9 x − 3 ⋅ x + 3 x + 3 = x → 9 lim ( x − 9 ) ( x + 3 ) x − 9 = x → 9 lim x + 3 1 = 9 + 3 1 = 6 1 Example 3: Find limit:
lim x → 1 x 2 − x x − 1 \mathop{\lim } \limits_{x \to 1} \frac{ {x^2} - \sqrt x }{\sqrt x - 1} x → 1 lim x − 1 x 2 − x Solution:
Here, it is helpful to do a substitution:
u = u ( x ) = x lim x → 1 u ( x ) = lim x → 1 x = 1
\begin{aligned}
&u = u(x) = \sqrt{x} \\
&\mathop {\lim }\limits_{x \to 1} u(x) = \mathop {\lim }\limits_{x \to 1} \sqrt x = 1
\end{aligned}
u = u ( x ) = x x → 1 lim u ( x ) = x → 1 lim x = 1 so that the limit expression can now be written as:
lim x → 1 x 2 − x x − 1 = lim u → 1 u 4 − u u − 1 = lim u → 1 u ( u 3 − 1 ) u − 1 = lim u → 1 u ( u − 1 ) ( u 3 + 1 u + 1 2 ) u − 1 = lim u → 1 u ( u 2 + u + 1 ) = 1 ( 1 + 1 + 1 ) = 3
\begin{aligned}
&\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - \sqrt x }}{{\sqrt x - 1}} = \mathop {\lim }\limits_{u \to 1} \frac{{{u^4} - u}}{{u - 1}} = \mathop {\lim }\limits_{u \to 1} \frac{{u({u^3} - 1)}}{{u - 1}} = \mathop {\lim }\limits_{u \to 1} \frac{{u(u - 1)({u^3} + 1u + {1^2})}}{{u - 1}} \\
&= \mathop {\lim }\limits_{u \to 1} u({u^2} + u + 1) = 1(1 + 1 + 1) = 3
\end{aligned}
x → 1 lim x − 1 x 2 − x = u → 1 lim u − 1 u 4 − u = u → 1 lim u − 1 u ( u 3 − 1 ) = u → 1 lim u − 1 u ( u − 1 ) ( u 3 + 1 u + 1 2 ) = u → 1 lim u ( u 2 + u + 1 ) = 1 ( 1 + 1 + 1 ) = 3
Example 4:
lim x → 1 x − 1 x 2 + 3 − 2 = lim x → 1 x − 1 x 2 + 3 − 2 ⋅ x 2 + 3 + 2 x 2 + 3 + 2 = lim x → 1 ( x − 1 ) ( x 2 + 3 + 2 ) x 2 + 3 − 4 = lim x → 1 ( x − 1 ) ( x 2 + 3 + 2 ) x 2 − 1 = lim x → 1 x 2 + 3 + 2 x + 1 = 4 2 = 2
\begin{aligned}
&\mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{\sqrt {{x^2} + 3} - 2}} = \\
&\mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{\sqrt {{x^2} + 3} - 2}} \cdot \frac{{\sqrt {{x^2} + 3} + 2}}{{\sqrt {{x^2} + 3} + 2}} = \\
&\mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)(\sqrt {{x^2} + 3} + 2)}}{{{x^2} + 3 - 4}} = \\
&\mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)(\sqrt {{x^2} + 3} + 2)}}{{{x^2} - 1}} = \\
&\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 3} + 2}}{{x + 1}} = \frac{4}{2} = 2
\end{aligned}
x → 1 lim x 2 + 3 − 2 x − 1 = x → 1 lim x 2 + 3 − 2 x − 1 ⋅ x 2 + 3 + 2 x 2 + 3 + 2 = x → 1 lim x 2 + 3 − 4 ( x − 1 ) ( x 2 + 3 + 2 ) = x → 1 lim x 2 − 1 ( x − 1 ) ( x 2 + 3 + 2 ) = x → 1 lim x + 1 x 2 + 3 + 2 = 2 4 = 2 Example 5: Find limit
lim x → 1 1 − x 2 − 3 x + 3 4 x 2 − 3 x − 1 \mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt {{x^2} - 3x + 3} }}{{\sqrt {4{x^2} - 3x - 1} }} x → 1 lim 4 x 2 − 3 x − 1 1 − x 2 − 3 x + 3 Solution:
lim x → 1 1 − x 2 − 3 x + 3 4 x 2 − 3 x − 1 = lim x → 1 1 − x 2 − 3 x + 3 4 x 2 − 3 x − 1 ⋅ 1 + x 2 − 3 x + 3 1 + x 2 − 3 x + 3 = = lim x → 1 1 − x 2 − 3 x + 3 ( 1 + x 2 − 3 x + 3 ) 4 x 2 − 3 x − 1 = lim x → 1 ( x − 1 ) ( − x − 4 ) ( 1 + x 2 − 3 x + 3 ) ( x − 1 ) ( 4 x − 1 ) = = lim x → 1 − x − 1 ( x + 4 ) ( 1 + x 2 − 3 x + 3 ) 4 x − 1 = lim x → 1 − 0 ⋅ 5 ( 1 + 1 ) 3 = 0
\begin{aligned}
&\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt {{x^2} - 3x + 3} }}{{\sqrt {4{x^2} - 3x - 1} }} = \mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt {{x^2} - 3x + 3} }}{{\sqrt {4{x^2} - 3x - 1} }} \cdot \frac{{1 + \sqrt {{x^2} - 3x + 3} }}{{1 + \sqrt {{x^2} - 3x + 3} }} = \\
&= \mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} - 3x + 3}}{{(1 + \sqrt {{x^2} - 3x + 3} )\sqrt {4{x^2} - 3x - 1} }} = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)( - x - 4)}}{{(1 + \sqrt {{x^2} - 3x + 3} )\sqrt {(x - 1)(4x - 1)} }} = \\
&=\mathop {\lim }\limits_{x \to 1} \frac{{ - \sqrt {x - 1} (x + 4)}}{{(1 + \sqrt {{x^2} - 3x + 3} )\sqrt {4x - 1} }} = \mathop {\lim }\limits_{x \to 1} \frac{{ - \sqrt 0 \cdot 5}}{{(1 + \sqrt 1 )\sqrt 3 }} = 0
\end{aligned}
x → 1 lim 4 x 2 − 3 x − 1 1 − x 2 − 3 x + 3 = x → 1 lim 4 x 2 − 3 x − 1 1 − x 2 − 3 x + 3 ⋅ 1 + x 2 − 3 x + 3 1 + x 2 − 3 x + 3 = = x → 1 lim ( 1 + x 2 − 3 x + 3 ) 4 x 2 − 3 x − 1 1 − x 2 − 3 x + 3 = x → 1 lim ( 1 + x 2 − 3 x + 3 ) ( x − 1 ) ( 4 x − 1 ) ( x − 1 ) ( − x − 4 ) = = x → 1 lim ( 1 + x 2 − 3 x + 3 ) 4 x − 1 − x − 1 ( x + 4 ) = x → 1 lim ( 1 + 1 ) 3 − 0 ⋅ 5 = 0
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