How Do You Further Simplify the Limit of a Square Root Expression?

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Homework Help Overview

The discussion revolves around simplifying the limit of a square root expression as h approaches 0, specifically the limit of the expression (√(a+h) - √a) / h. Participants are exploring the algebraic manipulation and implications of this limit in the context of calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the algebraic steps taken to simplify the limit, including multiplying by the conjugate. Questions arise about whether further simplification is necessary and what the limit evaluates to as h approaches 0.

Discussion Status

Some participants suggest that the expression does not need further simplification and point out the behavior of the limit as h approaches 0. Others provide insights into the relationship between the limit and differentiation, indicating a productive exploration of concepts related to calculus.

Contextual Notes

There is an underlying assumption that participants are familiar with limits and differentiation, which influences the discussion. Some posts hint at varying levels of understanding regarding these concepts.

pyrosilver
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Homework Statement



lim
[tex]_{}h\rightarrow0[/tex] ([tex]\sqrt{a+h}[/tex] - [tex]\sqrt{a}[/tex]) / h

3. ok so i multiplied by [tex]\sqrt{a+h}[/tex] + [tex]\sqrt{a}[/tex]. The top simplified to a+ h -a, aka h. the bottom is h([tex]\sqrt{a+h}[/tex] + [tex]\sqrt{a}[/tex]).

The h cancels, so I am left with

1 / ([tex]\sqrt{a+h}[/tex] + [tex]\sqrt{a}[/tex]

how do i simplify this further?

thanks!
 
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No need to simplify any further, just note what happens as [itex]h \to 0[/itex]. :wink:
 
Supposedly I can simplify it to

1
----
2[tex]\sqrt{a}[/tex]?
 
Which is what you get when you evaluate the limit as [itex]h \to 0[/itex].
 
It makes perfect sense as well that that is your answer. I am not sure of the extent of your understanding of differentiation, however, if you know enough, you should be able to recall that the derivative of a^(1/2) = 1/2a^(1/2) which is exactly what you got.
 
windwitch said:
It makes perfect sense as well that that is your answer. I am not sure of the extent of your understanding of differentiation, however, if you know enough, you should be able to recall that the derivative of a^(1/2) = 1/2a^(1/2) which is exactly what you got.

A clearer way of writing this would be a^(1/2) = 1/(2a^(1/2)). A literal reading of what you wrote would be 1/2*a^(1/2); that is, the fractional power would be in the numerator, which I'm sure you didn't intend.
 

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