Limit of a Function with Radicals in the Numerator

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SUMMARY

The limit of the function as h approaches 0 for the expression [√(5+h) - √(5-h)]/h simplifies to 2/√(5+h) + √(5-h). By multiplying the numerator and denominator by the conjugate, the limit is ultimately evaluated to √5/5. This solution effectively demonstrates the application of limits involving radicals and the importance of careful algebraic manipulation in calculus.

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Students studying calculus, particularly those focusing on limits and radical functions, as well as educators seeking to clarify concepts related to limit evaluation techniques.

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Homework Statement



Limit as h approaches 0 for [rad(5+h)-rad(5-h)]/h

Homework Equations


The Attempt at a Solution



limit as h approaches 0 for [(5+h)-(5-h)]/h[rad(5+h)+rad(5-h)]

limit as h approaches 0 for 2h/h[rad(5+h)+rad(5-h)]

limit as h approaches 0 for h/[rad(5+h)+rad(5-h)]

This was as far as I could get. Sorry if it's a little messy.
 
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Limit as h approaches 0 for [rad(5+h)-rad(5-h)]/h

That is Lim_(h to 0) [itex]\frac{\sqrt{5+h} - \sqrt{5-h}}{h}[/itex]

Multiply numerator and denominator by [itex]\sqrt{5+h} + \sqrt{5-h}[/itex]
When you work through steps, you obtain expression,...
[itex]\frac{2}{\sqrt{5+h} + \sqrt{5-h}}[/itex]

As h approaches 0, the expression approaches [itex]\frac{2}{\sqrt{5} + \sqrt{5}}[/itex]

Simplifying to [itex]\frac{\sqrt{5}}{5}[/itex]. DONE.

Note minor TEX/LATEX learning problems, "Lim as h approaches 0"
 
Thank you for clearing that up. Instead of cancelling the 'h', I canceled the 2 by mistake.
 

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