Hospital Rule
For the limit,
\ [\ Lim \ _ {Limits \ substack {X \ \\ to A (X \ to \ infty)}} \ FRAC {F (X)} {G (X)} \]
For type undetermined, i.e. $ \ FRAC {0} {0} $, \ (\ FRAC {\ infty} {\ infty} \) , \ (\ infty- \ infty \) , \ (0 \ CDOT \ infty \ ) , \ (0 ^ 0 \) , \ (0 ^ \ infty \) , \ (\ infty 0 \) ^ , \ (. 1 ^ \ infty \) and other forms of limit, can be directly or modified after use Hospitol method of the calculation.
Note: you can use the equivalent of endless laughter first with substitution after the calculation of equivalent infinitesimal
1. \ (\ FRAC {0} {0} \) type
Example One: Find the limit
\[\lim\limits_{\substack{x\to 0 } }\frac{x-tanx}{(sinx)^3}\]
解：代价无穷小代换：\(sinx\sim x\)
\[\lim\limits_{\substack{x\to 0 } }\frac{x-tanx}{(sinx)^3}\]
\[=\lim\limits_{\substack{x\to 0 } }\frac{x-tanx}{(x)^3}=\lim\limits_{\substack{x\to 0 } }\frac{(x-tanx)'}{(x^3)' }=\lim\limits_{\substack{x\to 0 } }\frac{1-sex^2x}{3x^2 } \]
\[=\lim\limits_{\substack{x\to 0 } }\frac{-tan^2x}{3x^2}=\lim\limits_{\substack{x\to 0 } }\frac{-x^2}{3x^2}\]
\[=-\frac{1}{3}\]
- $ tanx^2=secx^2-1$
2.\(tanx \sim x\)
Added: common equivalent infinitesimal \ (x \ to 0 \) when
\ [sinx \ sim x \]
\ [tanx \ sim x \]
\ [1-cosx \ sim \ frac {1} {2} x ^ 2 \]
\ [arcsinx \ sim x \]
\ [arctanx \ sim x \]
\ [a ^ x-1 \ sim xlina \]
\ [e ^ x-1 \ sim x \]
Dot: \ cdot
\ (A \ cdot b \)
Cross product: \ Times
\ (A \ Times b \)
Divided: \ div
\ (A \ div b \)
In Latex ~ represents a space, can be \ SIM escape instead of
\ [A \ sim B \]
$$A\sim B$$Example Two: Evaluating Limit
\[\lim\limits_{\substack{x \to 0 } } \frac{e^x+e^{-x}-2}{x-sinx}\]
\ [\ Color {# ea4335} {= \ lim \ limits _ {\ substack {x \ to 0}} \ frac {e ^ xe ^ {- x}} {1-cosx} ( first use Hospital Rule )} \]
\ [= \ Lim \ Limits _ {\ substack {X \ to 0}} \ FRAC {E ^ XE ^ {- X}} {\ FRAC {. 1} {2} X ^ 2} (equivalent infinitesimal :. 1-cosx \ SIM \ FRAC {. 1} {2} X ^ 2) \]
\ [= \ Lim \ Limits _ {\ substack {X \ to 0}} \ FRAC {E ^ X + E ^ {- X} } {x} (secondary use Hospital rule) \ {# Color 34a853} {\ rightarrow} \ {# 34a853 Color {} \ {2} FRAC {0}} \]
\ [= \ infty \]
2. $ \ frac {\ infty} {\ infty} $ type
Example Three: Find
\[\lim\limits_{\substack{x\to +\infty } }\frac{lnx}{x^{\alpha}},(\alpha>0)\]
解：\[=\lim\limits_{\substack{x\to +\infty } }\frac{ {1}\over{x}}{\alpha x^{\alpha-1}}=\lim\limits_{\substack{x\to +\infty } }\frac{1}{ \alpha\cdot x^{\alpha}}=0\]
\[(x\rightarrow +\infty,x^{\alpha}\to \infty )\]
3. \ (\ infty- \ infty \) type
Tips: Jane through differentiation, common denominator, rationalizing radical; variable substitution or the like into \ (\ frac {0} { 0} \) type, or \ (\ frac {\ infty} {\ infty} \) Aplastic calculation.
Example 4: seeking
\[\lim\limits_{\substack{x\to \frac{\pi}{2}} }(secx-tanx)\]
解：
\[\lim\limits_{\substack{x\to \frac{\pi}{2}} }(secx-tanx)\]
\[=\lim\limits_{\substack{x\to \frac{\pi}{2}} }(\frac{1}{cosx}-\frac{sinx}{cosx})\]
\[=\lim\limits_{\substack{x\to \frac{\pi}{2}} }(\frac{1-sinx}{cosx})\]
\[=\lim\limits_{\substack{x\to \frac{\pi}{2}} }\frac{cosx}{sinx} \to \frac{0}{1}\]
\[=0\]
4. \ (0 \ CDOT \ infty \) type
Converted to \ (\ frac {0} { 0} \) type, or \ (\ frac {\ infty} {\ infty} \) type
Example 5: seeking
\[\lim\limits_{\substack{x\to 0^+ } } sinxlnx\]
解：\[=\lim\limits_{\substack{x\to 0^+ } } \frac{lnx}{1\over{sinx}}\]
\[=\lim\limits_{\substack{x\to 0^+ } } \frac{lnx}{1\over{x}}=\lim\limits_{\substack{x\to 0^+ } } \frac{1\over{x}}{-{1\over{x^2}}}\]
\[=-\lim\limits_{\substack{x\to 0^+ } } x\]
\[=0\]
5. The \ (0 ^ 0, \ infty ^ 0,1 ^ {\ infty} \) type
Using \ (\ lim f (x) ^ {g (x)} = e ^ {\ lim g (x) lnf (x)} \) into \ (0 \ CDOT \ infty \) , then converted to \ ( \ frac {0} {0} \) type or \ (\ frac {\ infty} {\ infty} \) type
Example 6: To seek
\[\lim\limits_{\substack{x\to 0^+ } } x^{sinx}\]
解：\[\lim\limits_{\substack{x\to 0^+ } } x^{sinx}=e^{\lim\limits_{\substack{x\to 0^+ }}sinxln(x) }\]
\[=e^o=1\]
Example seven: seeking
\[\lim\limits_{\substack {x\to 0^+} } (1+cotx)^{{1}\over{lnx}}\]
解： \[=e^{\lim\limits_{\substack {x\to 0^+} } {{1}\over{lnx}}ln(1+cotx) }\]
\[=e^{\lim\limits_{\substack {x\to 0^+} } \frac{ln(1+cotx)}{lnx} }=e^{\lim\limits_{\substack {x\to 0^+} } \frac{ln(1+cotx)'}{(lnx)'} }\]
\[=e^{\lim\limits_{\substack {x\to 0^+} } {\left. {(\frac{-csc^2x }{1+cotx})} \middle / (\frac{1}{x})\right.} }\]
\[=e^{\lim\limits_{\substack {x\to 0^+} } {\left. {-x}\middle /{(sin^2x+sinxcosx)} \right.} }=e^{\lim\limits_{\substack {x\to 0^+} } {\left. {-x}\middle /{sinx(sinx+cosx)} \right.} }\]
\[=e^{\lim\limits_{\substack {x\to 0^+} } {\left. {-1}\middle /{(sinx+cosx)} \right.} }\]
\[=e^{-1}\]
