What function is being differentiated

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jwpaine

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What function is being differentiated

\(\displaystyle \L \lim_{h\to\0}\frac{\sqrt{1+h^{2}} -1}{h}\)

I know as h -> 0, the function approaches 0, because

\(\displaystyle \L \lim_{h\to\0}\frac{\sqrt{1+h^{2}} -1}{h}\cdot \frac{\sqrt{1+h^{2}} +1}{\sqrt{1+h^{2}} +1}\)

= \(\displaystyle \L \lim_{h\to\0}\, \frac{h^{2}}{h\sqrt{1+h^{2}} + h}\)

= \(\displaystyle \L \lim_{h\to\0}\, \frac{h}{\sqrt{h^{2}+ 1} + 1}\)

= \(\displaystyle \L \frac{0}{2} = 0\)

so I think that the function that is being differentiated is a constant function such as f(x) = c.... but what is the function?


Am I on the right path?
 
\(\displaystyle \L \lim_{h \rightarrow 0} \frac{\sqrt{1 + h^2 - 1}}{h} =\)

\(\displaystyle \L \lim_{h \rightarrow 0} \frac{\sqrt{h^2}}{h} =\)

\(\displaystyle \L \lim_{h \rightarrow 0} \frac{|h|}{h}\)

as h -> 0 from the left, |h|/h approaches -1

as h -> 0 from the right, |h|/h approaches 1

since the left limit does not equal the right limit, the limit does not exist.

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are you sure the limit wasn't ...

\(\displaystyle \L \lim_{h \rightarrow 0} \frac{\sqrt{1 + h^2} - 1}{h}\) ?
 
Hello, jwpaine!

Please check the original problem.

I'm sure there's a typo.
As skeeter pointed out, the function can be simplified first anyway.

What function is being differentiated?

\(\displaystyle \L \lim_{h\to\0}\frac{\sqrt{1+h^{2} -1}}{h}\)
Click to expand...

Besdies, that expression is not a difference quotient.

The numerator of the difference quotient is: \(\displaystyle \:f(x\,+\,h)\,-\,f(x)\)

I too considered: \(\displaystyle \L\:\lim_{h\to0}\frac{\sqrt{1+h^2}\,-\,1}{h}\)

. . but there's no function \(\displaystyle f(x)\) that produces that numerator.

 
As corrected, the function is \(\displaystyle f(x) = \sqrt {1 + x^2 }\) and
\(\displaystyle f'(0) = \lim _{h \to 0} \frac{{f(0 + h) - f(0)}}{h}\).
 
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