Limit at 0 + 0^+

Calculus Level pending

Consider the function f ( x ) = 1 x f(x)=\frac{1}{x} . For each positive value of h h , define θ h \theta_h to be the unique positive value that satisfies the equation f ( 2 + h ) = f ( 2 ) + h f ′ ( 2 + θ h h ) . f( 2 + h) = f(2) + h f'(2 + \theta_h h ). What is the value of lim ⁡ h → 0 + 1 θ h \displaystyle \lim_{h \to 0^+} \frac{1}{\theta_h} ?

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Mahindra Jain by Mahindra Jain

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1 solution

Tom Engelsman
Nov 8, 2020

Let us first solve for θ h \theta_{h} according to:

f ( 2 + h ) = f ( 2 ) + h f ′ ( 2 + θ h h ) ⇒ 1 2 + h = 1 2 − h ( 1 ( 2 + θ h h ) 2 ) f(2+h) = f(2) + hf'(2 +\theta_{h}h) \Rightarrow \frac{1}{2+h} = \frac{1}{2} - h(\frac{1}{(2+\theta_{h}h)^2}) ;

or θ h = 2 h + 4 − 2 h \theta_{h} = \frac{\sqrt{2h+4}-2}{h} . The limit in question is now calculated per:

lim ⁡ h → 0 + h 2 h + 4 − 2 ⇒ 1 1 / 2 h + 4 = 2 h + 4 = 2 \lim_{h \rightarrow 0^{+}} \frac{h}{\sqrt{2h+4}-2} \Rightarrow \frac{1}{1/\sqrt{2h+4}} = \sqrt{2h+4} = \boxed{2}

using L'Hoptial's Rule above.

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