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IIT-JEE 2009

LIMIT: IMPORTANT DEFINITIONS AND RESULTS - II

Author: anuraagchandra

Some expansions:
1. $\log_e(1+x)= x- \frac{x^2}{2}+\frac {x^3}{3}- \cdots to \quad \infty, -1 <x \underline <1$
2. $\log_e(1-x)=-x- \frac{x^2}{2}- \frac {x^3}{3}- \cdots to \quad \infty, -1 \underline <x < 1$
3. $e^x=1+\frac {x} {1!}+\frac {x^2} {2!}+\cdots to \quad \infty$
4. $e^{-x}=1-\frac {x}{1!}+\frac {x^2}{2!}+\cdots to \quad \infty$
5. $a^x = e ^{x\log_e a}$
6. $\sin x =x-\frac {x^3}{3!}+\frac {x^5}{5!}- \cdots to \quad \infty$
7. $\cos x=1-\frac {x^2}{2!}+\frac {x^4}{4!}- \cdots to \quad \infty$
8. $\tan x=x+\frac {x^3}{3}+\frac {2}{15}x^5+ \cdots to \quad \infty$
9. $\tan^{-1} x =x-\frac {x^3}{3}+\frac {x^5}{5}- \cdots$ , where $-1 \underline <x \underline <1$
Some Results on Limits:
- $\overset {}{\underset {f(x) \rightarrow \infty}{Lt}}\Big[ 1+\frac {k}{f(x)}\Big]^{f(x)}=e^k$ This is used only when both base and power are variables.
- $\overset {}{\underset {x \rightarrow a}{Lt}} \frac {x^n-a^n}{x-a}=na^{n-1}$ , where a is a constant
- General Form: $\overset {}{\underset {f(x)\rightarrow a}{Lt}}\frac {(f(x))^n-a^n}{f(x)-a}=na^{n-1}$
- $\overset {}{\underset {f(x)\rightarrow 0}{Lt}}\frac {\log e[1+f(x)]}{f(x)}=1$
- $\overset {}{\underset {f(x)\rightarrow 0}{Lt}}[1+kf(x)]^{1/f (x)}=e^k$ This is used only when both base and power are variables.
- $\overset {}{\underset {x\rightarrow a}{Lt}} [f(x)]^{g(x)}=\overset {}{\underset {x\rightarrow a}{Lt}}e^{g(x)}$ $^{[f(x)-1]}$ [ $1^{\infty}$ form] This can be used when indeterminate form is $1^{\infty}$
- $\overset {}{\underset {x \rightarrow 0}{Lt}}\cos x=1,$ $\overset {}{\underset {x \rightarrow 0}{Lt}}\frac {\sin x}{x}$ $=1, \overset {}{\underset {x\rightarrow 0}{Lt}}\frac {\tan x}{x}=1$ , where x is in radian
- $\overset {}{\underset {x\rightarrow 0}{Lt}}\frac {\sin^{-1}x}{x}$ $=1, \overset {}{\underset {x\rightarrow 0}{Lt}}\frac {\tan^{-1}x}{x}=1$
- Sandwich theorem: If
  
  Then $\overset {}{\underset {x \rightarrow a}{Lt}}g(x)$ â‰¤ $\overset {}{\underset {x \rightarrow a}{Lt}}f(x)$ â‰¤ $\overset {}{\underset {x \rightarrow a}{Lt}}h(x)$
  
  If $\overset {}{\underset {x \rightarrow a}{Lt}}g(x)=l$ and $\overset {}{\underset {x \rightarrow a}{Lt}}h(x)=l$ and , then $\overset {}{\underset {x \rightarrow a}{Lt}}f(x)=l$
Limit by changing into integrals:
Limit can be evaluated by changing into integral if
- $n \rightarrow \infty$
- given expression is the sum of a series in n.
- each term of the series tends to zero as n tends to infinity.
- Working Rule:
  Write the given limit as $\overset {}{\underset {n \rightarrow \infty}{Lt}}$ $\overset {bn+\beta}{\underset {an+\alpha}{\sum}}\frac {1}{n}f \Big( \frac {r}{n} \Big)$
  Finally, put $\overset {}{\underset {n \rightarrow \infty}{Lt}}$ $\overset {bn+\beta}{\underset {an+\alpha}{\sum}}\frac {1}{n}f \Big( \frac {r}{n} \Big)$ $=\int_{a}^{b}f(x)dx$
  For this put dx in place of x, x in place of $\frac {r}{n}, \int_{}^{}$ in place of $\sum$ and
Lower limit integral
Upper limit of integral