2

find the limit $$\lim_{x\to\pi/4}\frac{\tan(x)-1}{x-\pi/4}$$

direct substitution results in $0/0$

and it seems that there's no way to factor it

When I look at its graph

It's clear that it has the limit 2 as $x$ approaches $\pi/4$

the question is How to factor this function ?? !

user2097
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Maher
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6 Answers6

4

Without applying L'Hospital's Rule, $$\eqalign{ & \tan (x - {\pi \over 4}) = {{\tan x - 1} \over {1 + \tan x}} \longrightarrow \tan x - 1 = (1 + \tan x)\tan (x - {\pi \over 4}) \cr & \mathop {\lim }\limits_{x \to {\pi \over 4}} {{\tan x - 1} \over {x - {\pi \over 4}}} \cr & = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{(1 + \tan x)\tan (x - {\pi \over 4})} \over {x - {\pi \over 4}}} \cr & = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{(1 + \tan x)(x - {\pi \over 4})} \over {x - {\pi \over 4}}} \cr & = \mathop {\lim }\limits_{x \to {\pi \over 4}} (1 + \tan x) \cr & = 2 \cr} $$

Shine Mic
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  • At the end of the day the problem comes down to $\lim_{x \to 0} \frac{\tan(x)}{x} = 1$. – Ian Oct 17 '14 at 15:01
  • @Ian Of course. The most common approach is equivalent infinitesimal replacement when solving the limits. – Shine Mic Oct 17 '14 at 15:05
3

Use L'Hopital's rule. $\lim_{x\rightarrow \frac{\pi}{4}} \frac{\tan{x}-1}{x-\frac{\pi}{4}}=\lim_{x\rightarrow \frac{\pi}{4}}\frac{1+\tan^2(x)}{1}=2$

CLAUDE
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3

By using Taylor's expansion we have $\tan(x)\simeq 1+2(x-\frac{\pi}{4})$, around $x=\frac{\pi}{4}$, hence we can say:

$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\tan(x)-1}{x-\frac{\pi}{4}}=\lim_{x\rightarrow \frac{\pi}{4}}\frac{1+2(x-\frac{\pi}{4})-1}{x-\frac{\pi}{4}}=2$

CLAUDE
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  • I know nothing about Taylor's expansion :( . no other ways of solving this problem other than Taylor's expansion or L'Hopital's rule..? – Maher Oct 15 '14 at 21:50
  • No, I think you should practice more to become familiar with these basic concepts of calculus. – CLAUDE Oct 15 '14 at 21:51
1

Because $\tan(\frac{\pi}{4})=1$ this can be thinked as difference quotient. $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\tan\left(x\right)-1}{x-\frac{\pi}{4}}$ is just like $\lim_{x\rightarrow x_0}\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}$ where $f(x)$ is $\tan(x)$ and $x_0$ is $\frac{\pi}{4}$. Now, $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\tan\left(x\right)-1}{x-\frac{\pi}{4}}=f'(x)$. $f'(x)=D(\tan x)=1+\tan^2x$. And then $f'(\frac{\pi}{4})=1+\tan^2(\frac{\pi}{4})=1+1^2=2$

José Carlos Santos
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KiviKnuuti
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1

With a shift of the variable ($t\to0$),

$$\frac{\tan(t+\frac\pi4)-1}{t}=\frac{\dfrac{1+\tan t}{1-\tan t}-1}t=\frac{2\tan t}{t(1-\tan t)}\to2.$$

0

It is not necessary to factor! This limit is not so coincidentally the derivative of $f(x)=\tan(x)$ at the point $x=\pi/4$. By a substitution it can be shown $$ f'(a) = \lim_{x \to a }\frac{f(x) - f(a) }{x-a} $$

Specifically $$ \tan'(\pi/4) =\lim_{x\to\pi/4}\frac{\tan(x)-\tan(\pi/4)}{x-\pi/4}$$ Then using the fact that $ \tan (x) ' =\sec^2( x) $ we immediately get the limit is $$ \sec\left(\pi/4\right)^2 = \left(\frac {2}{\sqrt 2 } \right)^2 = 2 $$

john
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