Full solution

Q. Find

\lim _{h \rightarrow 0} \frac{2 \tan \left(\frac{\pi}{3}+h\right)-2 \tan \left(\frac{\pi}{3}\right)}{h}

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Choose

1

answer:

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(A)

1

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(B)

2

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(C)

8

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(D) The limit doesn't exist

Identify Problem: Identify the limit problem and recognize it as a derivative problem at a specific point.
Use Derivative Definition: Use the definition of the derivative $f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ for the function $f(x) = 2\tan(x)$ at $x = \frac{\pi}{3}$ .
Plug in Values: Plug in the values into the derivative formula: $f'(\frac{\pi}{3}) = \lim_{h \to 0}\left(\frac{2\tan(\frac{\pi}{3}+h)-2\tan(\frac{\pi}{3})}{h}\right).$
Simplify Expression: Simplify the expression inside the limit: $2\tan(\frac{\pi}{3}+h)-2\tan(\frac{\pi}{3}) = 2(\tan(\frac{\pi}{3}+h)-\tan(\frac{\pi}{3}))$ .
Apply Tangent Formula: Apply the tangent addition formula: $\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$ , where $a = \frac{\pi}{3}$ and $b = h$ .
Substitute Values: Substitute $\tan(\frac{\pi}{3}) = \sqrt{3}$ into the formula: $\tan(\frac{\pi}{3}+h) = \frac{\sqrt{3}+\tan(h)}{1-\sqrt{3}\tan(h)}$ .
Plug Expression Back: Plug the expression for $\tan(\frac{\pi}{3}+h)$ back into the limit: $\lim_{h \to 0}\left(\frac{2\left(\sqrt{3}+\tan(h)\right)}{1-\sqrt{3}\tan(h)}-\sqrt{3}\right)/h$ .
Simplify Inside Limit: Simplify the expression inside the limit: $\frac{2(\sqrt{3}+\tan(h))}{1-\sqrt{3}\tan(h)}-2\sqrt{3}$ /h = $2\left(\frac{\tan(h)}{1-\sqrt{3}\tan(h)}\right)/h$ .
Divide by h: Divide by h inside the limit: $\lim_{h \to 0}\left(\frac{2\tan(h)}{(1-\sqrt{3}\tan(h))h}\right)$ .
Recognize Approach: Recognize that as $h$ approaches $0$ , $\frac{\tan(h)}{h}$ approaches $1$ .
Substitute in Limit: Substitute $\tan(h)/h$ with $1$ in the limit: $\lim_{h \to 0}\left(\frac{2}{1-\sqrt{3}\tan(h)}\right)$ .
Evaluate Limit: Evaluate the limit as $h$ approaches $0$ : $\frac{2}{1-\sqrt{3}\tan(0)} = \frac{2}{1-0} = 2$ .
Conclude Solution: Conclude that the value of the limit is $2$ , which corresponds to answer choice $(B)$ .