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cestrela7 [59]
2 years ago
6

Im h--> 0 ((Sqrt 9+h)-3)/h

Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
6 0
\lim_{h \to 0}  \frac{ \sqrt{9+h}-3}{h} = \frac{ \sqrt{9+0} -3}{0}= \frac{0}{0} \\  \lim_{h \to 0}  \frac{ \sqrt{9+h}-3}{h}* \frac{ \sqrt{9+h}+3 }{ \sqrt{9+h} +3}= \\ = \lim_{h \to 0}   \frac{9+h-9}{h( \sqrt{9+h} +3)}  = \\   \lim_{n \to 0} \frac{h}{h( \sqrt{9+h}+3) }= \\ = \frac{1}{ \sqrt{9} +3}= 
= 1/6
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Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

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