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May 12th, 2015, 02:32 PM   #1
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lim (1 + tan x)^(2/x)
x → 0

Last edited by skipjack; May 12th, 2015 at 10:51 PM.
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May 12th, 2015, 02:53 PM   #2
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$\displaystyle \lim_{x\to0}\left(1+\tan x\right)^{2/x}=\lim_{x\to0}\left[\left(1+\tan x\right)^{1/x}\right]^2=e^2$ as $\displaystyle \lim_{x\to0}\frac{\tan x}{x}=1$
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May 12th, 2015, 09:08 PM   #3
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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., higher-order 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 higher-order 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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