Solving One-Sided Limit: lim x→4⁻√x-2/x-4

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Homework Help Overview

The problem involves evaluating the one-sided limit as x approaches 4 from the left for the expression (√x - 2) / (x - 4). Participants are exploring methods to resolve the indeterminate form encountered at this limit.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss substituting values close to 4 to evaluate the limit and note the resulting indeterminate form of 0/0. Some suggest using the conjugate to simplify the expression, while others introduce the concept of epsilon to analyze the limit more rigorously.

Discussion Status

The discussion is active, with participants sharing different approaches to tackle the limit. There is no explicit consensus on a single method, but various strategies are being explored, including factoring and using conjugates.

Contextual Notes

Participants note that the function is not defined at x = 4, which contributes to the indeterminate form. There is an emphasis on the necessity of careful evaluation when approaching limits involving square roots.

JennyInTheSky
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Homework Statement


lim_{x\rightarrow4^{-}} \frac{\sqrt{x}-2}{x-4}


Homework Equations


Typical methods used in solving one-sided limit.


The Attempt at a Solution


I plug in something a little bit smaller than four, like 3.999999 into x, and I get \frac{something a little less than zero}{something a little less than zero} to equal 1. But the answer is 1/4. How is this so?
 
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JennyInTheSky said:

Homework Statement


lim_{x\rightarrow4^{-}} \frac{\sqrt{x}-2}{x-4}


Homework Equations


Typical methods used in solving one-sided limit.


The Attempt at a Solution


I plug in something a little bit smaller than four, like 3.999999 into x, and I get \frac{something a little less than zero}{something a little less than zero} to equal 1. But the answer is 1/4. How is this so?
But see that the function at x=4 is not defined, right? so if you just keep plugging nr. you will get zero over zero, what i would do here, is multiply top and bottom by the conjugate of your numerator, this way things will clean up nicely.

Then, you can keep plugging nrs whicha re very very close to 4 from the left, in order to determine the sign. What i alway do is i introduce an e>0, e-->0 (e stands for epsilon) and so i let then x-->4-e, but that's just me.

Because sometimes it is not sufficient to take numbers close to, say 4 in this case. etc
 
sutupidmath said:
But see that the function at x=4 is not defined, right? so if you just keep plugging nr. you will get zero over zero, what i would do here, is multiply top and bottom by the conjugate of your numerator, this way things will clean up nicely.

Then, you can keep plugging nrs whicha re very very close to 4 from the left, in order to determine the sign. What i alway do is i introduce an e>0, e-->0 (e stands for epsilon) and so i let then x-->4-e, but that's just me.

Because sometimes it is not sufficient to take numbers close to, say 4 in this case. etc

You could just factorise the bottom x-4=(\sqrt{x}-2)(\sqrt{x}+2)
 
Focus said:
You could just factorise the bottom x-4=(\sqrt{x}-2)(\sqrt{x}+2)

Either way!...lol... whatever works for you...
 

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