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I need help with this limit: $$ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $$

I've tried using the $$ \tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b} $$ with $(x+a)$ as $a$ and $(x)$ as $b$ but having difficulty getting beyond that.

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  • $\begingroup$ what have you tried so far? $\endgroup$ – Donlans Donlans Sep 5 at 0:00
  • $\begingroup$ @DonlansDonlans I've tried using the $$ tan(a-b) = \frac{tan(a)-tan(b)}{1+tan(a)tan(b)} $$ with (x+a) as a an (x) as b but having difficulty getting beyond that. $\endgroup$ – David Budiarto Sep 5 at 0:07
  • $\begingroup$ Try using the angle-addition formula to break up $\tan(x+a)$. Then put everything over a common denominator (which I expect will be $\cos x \cos a$) and see whether that helps. $\endgroup$ – Robert Shore Sep 5 at 0:17
  • $\begingroup$ you will also need to remember the special limit $\lim_{x\to\ 0} \frac{\sin x}{x} = 1$ $\endgroup$ – Donlans Donlans Sep 5 at 0:28
  • $\begingroup$ There are 2 steps to doing any trig limit. Using an angle addition formula and then remembering $\lim_{x \to 0} \frac{\sin x}{x} = 1$. $\endgroup$ – Charles Hudgins Sep 5 at 0:31
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The best way to do this problem is to realize that the limit is, by definition, $\tan ' (x)$. Knowing that $\tan x = \frac{\sin x}{\cos x}$ and knowing $\sin '(x) = \cos x$ and $\cos ' (x) = -\sin x$, you can calculate the limit with the quotient rule.

As you've said in a comment, the key to doing this problem straight ahead without first computing $\sin'(x)$ and $\cos '(x)$ is to use the identity $$ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} $$ Substituting, we have \begin{align} \frac{\tan(x + a) - \tan x}{a} &= \frac{\frac{\tan x + \tan a}{1 - \tan x \tan a} - \tan x}{a} \\&=\frac{\tan x + \tan a - (1 - \tan x \tan a) \tan x}{a(1 - \tan x \tan a)} \\&= \frac{\tan x + \tan a(1 + \tan^2 x) - \tan x}{a - a\tan x \tan a} \\&= \frac{\tan a \sec^2 x}{a - a \tan x \tan a} \\&= \frac{\tan a}{a} \cdot \frac{1}{1 - \tan x \tan a} \cdot \sec^2 x \end{align} So, if we can just show $\lim_{a \to 0} \frac{\tan a}{a} = 1$ and $\lim_{a \to 0} \frac{1}{1 - \tan x \tan a} = 1$, we'll have arrived at the expected result.

The second limit is easy. $$ \lim_{a \to 0} \frac{1}{1 - \tan x \tan a} = \frac{1}{1 - \tan x \lim_{a \to 0} \tan a} = \frac{1}{1 - \tan x \cdot 0} = 1 $$ The first limit is a bit trickier. On geometric grounds, one can argue that $$ \cos a \leq \frac{\sin a}{a} \leq 1 $$ Hence $$ 1 \leq \frac{\tan a}{a} \leq \sec a $$ Since $\lim_{a \to 0} \sec a = 1$, the squeeze theorem tells us that $\lim_{a \to 0} \frac{\tan a}{a} = 1$.

Now all we need to do is pass the limit on the difference quotient calculation above, use the two limits we just "proved" (a more careful proof might be called for depending on the demands of the course), and conclude that the limit is $\sec^2 x$.

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  • $\begingroup$ Stare at this picture to convince yourself of the geometric inequality asserted above. Hint: try to prove the reverse inequality where each term is inverted. $\endgroup$ – Charles Hudgins Sep 5 at 0:29
  • $\begingroup$ Can you elaborate further on the substitution part? $\endgroup$ – David Budiarto Sep 5 at 1:29
  • $\begingroup$ I'll add a step. Did that clear things up? $\endgroup$ – Charles Hudgins Sep 5 at 1:45
  • $\begingroup$ Yes, it did, thanks! $\endgroup$ – David Budiarto Sep 5 at 3:03
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$$\lim_{a \to 0} \frac{\tan(x+a)- \tan x}{a}= \lim_{a \to 0} \frac{\frac{\sin(x+a)}{\cos(x+a)}- \frac{\sin x}{\cos x}}{a}\\ =\lim_{a \to 0} \frac{\cos x \sin(x+a)-\sin x \cos (x+a)}{a(\cos x \cos (x+a))}\\ = \lim_{a \to 0} \frac{\cos x (\sin x \cos a + \cos x \sin a) - \sin x (\cos x \cos a - \sin x \sin a)}{a\cos x \cos (x+a)}\\ = \lim_{a \to 0} \frac{\sin a (\cos^2 x+ \sin^2 x)}{a \cos x \cos (x+a)}= \lim_{a \to 0} \frac{\sin a}{a} \frac {1}{\cos x \cos (x+a)}= \frac {1}{\cos^2 x}= \sec^2 x.$$

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The limit fits the definition of derivative of $\tan x$.

If you know the answer you just get

$$\lim_{a \to 0} =\frac{\tan(x+a)-\tan(x)}{a} = \frac {d}{dx} (\tan x) = \sec ^2 x$$

If you do not know the answer right away you try to use the quotient rule on $\frac {\sin x}{\cos x}$ to get the result.

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$\begin{align} {\tan(x+a) - \tan(x) \over a} &={1\over a}\left({\tan(x) + \tan(a) \over 1-\tan(x)\tan(a)} - \tan(x)\right) \cr &= {\tan(x)+\tan(a) - \tan(x)(1-\tan(x)\tan(a))\over a(1-\tan(x)\tan(a))} \cr &= \left({\tan a \over a}\right) \left( {1 + \tan^2(x) \over 1-\tan(x)\tan(a)} \right) \end{align}$

I use $\tan(a)$ taylor series for the limit: $\tan(a) = a + O(a^3)$

$\begin{align} \displaystyle{\lim_{a \to 0} {\tan(x+a) - \tan(x) \over a}} &= \displaystyle{\left(\lim_{a \to 0}{\tan a \over a}\right)} \left(\displaystyle{\lim_{a \to 0}{1 + \tan^2(x) \over 1-\tan(x)\tan(a)}} \right) \cr &= \displaystyle{\left(\lim_{a \to 0}{a \over a}\right)} \left(\displaystyle{\lim_{a \to 0}{1 + \tan^2(x) \over 1-a\;\tan(x)}} \right) \cr &= (1)\left({\sec^2(x) \over 1-0} \right) \cr &= \sec^2(x) \end{align}$

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