May 12th, 2015, 02:32 PM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  limit
lim (1 + tan x)^(2/x) x → 0 Last edited by skipjack; May 12th, 2015 at 10:51 PM. 
May 12th, 2015, 02:53 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,823 Thanks: 1049 Math Focus: Elementary mathematics and beyond 
$\displaystyle \lim_{x\to0}\left(1+\tan x\right)^{2/x}=\lim_{x\to0}\left[\left(1+\tan x\right)^{1/x}\right]^2=e^2$ as $\displaystyle \lim_{x\to0}\frac{\tan x}{x}=1$

May 12th, 2015, 09:08 PM  #3 
Member Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 
When $x\to0$ (or any finite number, really), Taylor series is a good way to solve limit problems. It converts less familiar functions like the trigonometrics into their more familiar polynomial counterparts. When you see tangent, in this case, as a polynomial, you can focus on the dominant term and ignore the rest (i.e., higherorder terms or H.O.T.). After taking the logarithm of the limit, we get $\displaystyle\ln y=\frac{2}{x}\ln\left(1+\tan x\right)$ The Taylor series for $\tan x$ can be derived from $\displaystyle\frac{\sin x}{\cos x}=\frac{x\frac{x^3}{3!}+\ldots}{1\frac{x^2}{2!}+\ldots}$. After doing the long division, you obtain $\displaystyle\tan x=x+\frac{x^3}{3}+\ldots$ The dominant term in a polynomial when $x\to0$ is the lowest exponent of $x$, which in the case of tangent is the $x^1$ term. Substituting that in for $\tan x$, we get $\displaystyle\ln y=\frac{2}{x}\ln\left(1+[x+\ldots]\right)$ Next, the Taylor series of $\ln(1+x)$ for $x\lt1$ equals $\displaystyle x\frac{x^2}{2}+\frac{x^3}{3}\ldots$ Extrapolating from that, we replace the $\ln(1+[x+\ldots])$ with $\displaystyle\ln y=\frac{2}{x}\left(x\ldots\right)$ At this point, it's evident that multiplying the $(x\ldots)$ by $\displaystyle\frac{2}{x}$ yields the new dominant term of $2$, since all of the higherorder terms have their $x's$ with an exponent $\geq1$. If $\displaystyle\ln y=2\ldots$, then $y=e^{2\text{H.O.T.}}$ In the limit, $x\to0$, all of the H.O.T. with $x's$ disappear, and the answer is simply $e^2$. 

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