Limits Problem Homework: Solving for x in (x^2-5x+4) / (sin(x^1/2) - 2)

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Homework Help Overview

The problem involves evaluating the limit of the expression (x^2-5x+4) / (sin(√x) - 2) as x approaches 4. The context is within the subject area of calculus, specifically limits and continuity.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to factor the numerator and notes that both the numerator and denominator approach zero when substituting x=4. Some participants question the applicability of L'Hôpital's rule, while others clarify that it cannot be used due to a lack of prior learning on the topic. One participant suggests a substitution to simplify the limit evaluation.

Discussion Status

The discussion is active, with participants exploring different approaches to the limit problem. There is a suggestion of a substitution method, and some participants are considering how to proceed without L'Hôpital's rule. No consensus has been reached yet.

Contextual Notes

Participants note that L'Hôpital's rule is not an option due to the current curriculum constraints. The discussion also reflects uncertainty about the next steps after identifying the indeterminate form.

ganondorf29
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Homework Statement



lim (x^2-5x+4) / ((sin(x^1/2) - 2))
x->4



Homework Equations





The Attempt at a Solution



Well I factored the top out to be (x-1)(x-4) and if I plug in 4 I get 0 in the numerator. In the denominator, if I plug in 4, I also end up with zero. I am not too sure what I am supposed to do now
 
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[tex]\lim_{x \rightarrow 4}\frac{x^2-5x+4}{sin(\sqrt x-2)}[/tex]

Are u allowed to apply l'hospital's rule? If so, then it will work nicely!
 
No, we can't use l'hospital's rule because we haven't learned it yet.
 
ganondorf29 said:
No, we can't use l'hospital's rule because we haven't learned it yet.

Ok, then here it is what just popped into my head,

i would let [tex]\sqrt x-2=t[/tex] so when x-->4, t-->0

now

[tex]\lim_{x \rightarrow 4}\frac{x^2-5x+4}{sin(\sqrt x-2)}=\lim_{x \rightarrow 4}\frac{(x-1)(\sqrt x-2)(\sqrt x+2)}{sin(\sqrt x-2)}=\lim_{t\rightarrow 0}\frac{t}{sin(t)}(t+4)[(t+2)^2-1][/tex]

now:

[tex]\lim_{t\rightarrow 0}\frac{t}{sint}=\lim_{t\rightarrow 0}{\frac{1}{\frac{sint}{t}}=1[/tex]

I think the rest is easy!
 

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