Methods of finding various limits

 

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Direct substitution

• $\underset{x->3}{\lim }(x+1)$

x The limit of is close to 3, Is in the 3 near , Let's go straight to x=3 Plug in x+1 Middle computation , have to 4.

There are some exceptions ：

• $\frac{oftenCount}{\infty }=0$
• $\frac{\infty }{oftenCount}=\infty$
• $\frac{NotzerooftenCount}{0}=\infty$
• ${\infty }^{>0}=\infty$
• ${\infty }^{<0}=\frac{1}{{\infty }^{>0}}=0$
• $n^{\infty }=0,0>n>1$
• $n^{\infty }=\infty ,n>1$

 

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$\frac{\infty }{\infty }$ type

Some problems cannot be solved directly , such as $\frac{\infty }{\infty }$ type , It's not a specific number , It's a trend .

• $\underset{x->\infty }{\lim }\frac{x^{100}+x^{-1001}+x}{x^{1000}+2x}$

 

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solution ： Grasp the main trends

In many trends （∞） in , We need to find the biggest trend , Because that is the most influential item .

$\frac{\infty }{\infty }$ type , Solving steps ：

• Identify trends
• Look at the index , molecular 、 The denominator retains the largest trend

$\underset{x->\infty }{\lim }\frac{x^{100}+x^{-1001}+x}{x^{1000}+2x}$

• = $\underset{x->\infty }{\lim }\frac{{\infty }^{100}+{\infty }^{-1001}+\infty }{{\infty }^{1000}+2\infty }$
• = $\underset{x->\infty }{\lim }\frac{\infty +0+\infty }{\infty +\infty }$
• = $\underset{x->\infty }{\lim }\frac{x^{100}}{x^{1000}}$
• = $\underset{x->\infty }{\lim }\frac{1}{x^{900}}$
• = $\underset{x->\infty }{\lim }\frac{1}{{\infty }^{900}}$
• = $\frac{1}{\infty }$
• = $0$
   

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solution ： Use lobita's law

 

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$\frac{0}{0}$ type

$\underset{x->0}{\lim }\frac{x}{sinx}=\frac{0}{0}$

When put $x->0$ After substituting into the formula , Will become $\frac{0}{0}$, There will also be no solution .

 

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solution ： Replace with equivalent infinitesimal

When a part tends to 0 when , There are five situations ：

Case one ,$x->0,sinx=x$,

• $\underset{x->0}{\lim }\frac{x}{sinx}=\frac{x}{x}=1$

The second case ,$1-cos\Delta$ Variable to $\frac{1}{2}{\Delta }^2$

• $\underset{x->0}{\lim }\frac{1-cosx}{x}=\underset{x->0}{\lim }\frac{\frac{1}{2}{x}^2}{x}$

The nature of the following three cases is the same , Are all forms of substitution .

 

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solution ： Use lobita's law

If the unknown $x->0、x->\infty$ After substitution , The formula is $\frac{0}{0}$ or $\frac{\infty }{\infty }$ , be $\lim \frac{f(x)}{g(x)}=\lim \frac{f^{\prime }(x)}{g^{\prime }(x)}$, Molecules become derivatives of molecules 、 The denominator becomes the derivative of the denominator .
 

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$\infty \cdot 0$ type

$\underset{x->\infty }{\lim }x(cos\frac{1}{x}-1)$

• = ∞ · (cos 0 - 1)
• = ∞ · 0

Direct substitution encounters $\infty \cdot 0$, There is no result .

We have another solution ：

• Find the simplest one a
• Turn this item into $\frac{1}{\frac{1}{a}}$

$\underset{x->\infty }{\lim }x(cos\frac{1}{x}-1)$

• = $\underset{x->\infty }{\lim }\frac{1}{\frac{1}{x}}(cos\frac{1}{x}-1)$
• = $\underset{x->\infty }{\lim }\frac{cos\frac{1}{x}-1}{\frac{1}{x}}$
• = $\frac{0}{0}$
   

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Index 、 There are all bases x The limits of

Form like ：$\underset{x->0}{\lim }(1+3x{)}^{\frac{2}{sinx}}$

hold $At\; the\; end\; of{Count}^{fingerCount}$ become $e^{fingerCount\cdot lnAt\; the\; end\; ofCount}$

$\underset{x->0}{\lim }(1+3x{)}^{\frac{2}{sinx}}=\underset{x->0}{\lim }{e}^{\frac{2}{sinx}ln(1+3x)}$

= $\underset{x->0}{\lim }{e}^{\frac{2ln(1+3x)}{sinx}}$

$\underset{x->?}{\lim }{e}^{fingerCount}=e^{\underset{x->?}{\lim }fingerCount}$

= $e^{\underset{x->0}{\lim }}\frac{2ln(1+3x)}{sinx}$
 

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Left and right limits of function

It is necessary to find the left and right limits

There are three limits , Only through the most primitive method — Find the left and right limits .

• The first category , Functions are segmented functions with braces , The required limit is the limit at the segment point .

• The second category , Count $g^{(x)}$ stay $g(x)$ The denominator of is 0 Limit at .

• The third category ,$arctang(x)$ stay $g(x)$ The denominator of is 0 Limit at

How to do questions ：

• First find the left limit 、 Right limit
• When the left limit = Right limit = Not for ∞ The number of hours , Functional limits exist , And limit = Left limit = Right limit
• When the left limit = Right limit = -∞ perhaps +∞ when , The limit of the function is ∞ / non-existent / There are no limits
• When the left limit != Right limit And Existence is not for ∞ The value of , The function limit does not exist And Not for ∞