Class 11: Limits – Lecture Notes

$\displaystyle \textbf{1. Existence of a Limit}$
$\displaystyle \text{The limit }\lim_{x\to a}f(x)\text{ exists }\Leftrightarrow \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$
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$\displaystyle \textbf{2. Limit and Value of a Function}$
$\displaystyle \text{For a function }f(x)\text{ and a real number }a,\;\lim_{x\to a}f(x)\text{ and }f(a)\text{ may not be same.}$
$\displaystyle \text{In fact:}$
$\displaystyle \text{(i) }\lim_{x\to a}f(x)\text{ exists but }f(a)\text{ (the value of }f(x)\text{ at }x=a\text{) may not exist.}$
$\displaystyle \text{(ii) The value }f(a)\text{ exists but }\lim_{x\to a}f(x)\text{ does not exist.}$
$\displaystyle \text{(iii) }\lim_{x\to a}f(x)\text{ and }f(a)\text{ both exist but are unequal.}$
$\displaystyle \text{(iv) }\lim_{x\to a}f(x)\text{ and }f(a)\text{ both exist and are equal.}$
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$\displaystyle \textbf{3. Laws of Limits}$
$\displaystyle \text{Let }\lim_{x\to a}f(x)=l\text{ and }\lim_{x\to a}g(x)=m.\text{ If }l\text{ and }m\text{ both exist, then}$
$\displaystyle \text{(i) }\lim_{x\to a}kf(x)=k\lim_{x\to a}f(x)$
$\displaystyle \text{(ii) }\lim_{x\to a}(f\pm g)(x)=\lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)=l\pm m$
$\displaystyle \text{(iii) }\lim_{x\to a}(fg)(x)=\lim_{x\to a}f(x)\lim_{x\to a}g(x)=lm$
$\displaystyle \text{(iv) }\lim_{x\to a}\left(\frac{f}{g}\right)(x)=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}=\frac{l}{m}$
$\displaystyle \text{(v) }\lim_{x\to a}\{f(a)\}^{g(x)}=l^m$
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$\displaystyle \textbf{4. Standard Limits}$
$\displaystyle \text{Following are some standard limits:}$
$\displaystyle \text{(i) }\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$
$\displaystyle \text{(ii) }\lim_{x\to0}\frac{\sin x}{x}=1$
$\displaystyle \text{(iii) }\lim_{x\to0}\frac{\tan x}{x}=1$
$\displaystyle \text{(iv) }\lim_{x\to a}\frac{\sin(x-a)}{x-a}=1$
$\displaystyle \text{(v) }\lim_{x\to a}\frac{\tan(x-a)}{x-a}=1$
$\displaystyle \text{(vi) }\lim_{x\to0}\frac{\log(1+x)}{x}=1$
$\displaystyle \text{(vii) }\lim_{x\to0}\frac{a^x-1}{x}=\log_e a,\;a\neq0,\;a>1$
$\displaystyle \text{(viii) }\lim_{x\to0}\frac{e^x-1}{x}=1$
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