Evaluating $ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $

Question

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.

Answer 1

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 $\sec^2 x$.

Answer 2

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.

Answer 3

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

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

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

Answer 4

$$\lim_{a \to 0} \frac{\tan(x+a)- \tan x}{a}= \lim_{a \to 0} \frac{\frac{\sin(x+a)}{\cos(x+a)}- \frac{\sin x}{\cos x}}{a}\\ =\lim_{a \to 0} \frac{\cos x \sin(x+a)-\sin x \cos (x+a)}{a(\cos x \cos (x+a))}\\ = \lim_{a \to 0} \frac{\cos x (\sin x \cos a + \cos x \sin a) - \sin x (\cos x \cos a - \sin x \sin a)}{a\cos x \cos (x+a)}\\ = \lim_{a \to 0} \frac{\sin a (\cos^2 x+ \sin^2 x)}{a \cos x \cos (x+a)}= \lim_{a \to 0} \frac{\sin a}{a} \frac {1}{\cos x \cos (x+a)}= \frac {1}{\cos^2 x}= \sec^2 x.$$