Using the epsilon and delta definition to prove limit

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SUMMARY

The limit of the function $\sqrt{x}$ as x approaches 9 is definitively 3. To prove this using the epsilon-delta definition, one can substitute $x=t^2$, transforming the limit to $\lim_{t\to3}t=3$. This method effectively demonstrates the limit's behavior, confirming that the original limit is positive and necessitating the condition $|t|=t$.

PREREQUISITES
  • Epsilon-delta definition of limits
  • Understanding of square root functions
  • Basic algebraic manipulation
  • Knowledge of limit notation and properties
NEXT STEPS
  • Study the epsilon-delta definition in depth
  • Practice proving limits of nonlinear functions
  • Explore transformations in limit proofs, such as $x=t^2$
  • Learn about continuity and its implications for limits
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Students in calculus, mathematics educators, and anyone looking to deepen their understanding of limit proofs using the epsilon-delta definition.

cbarker1
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Find the limit L. Then use the epsilon-delta definition to prove that the limit is L.

$\sqrt(x)$ as x approaches 9

I figure out the first part of the question. the Answer is three. Yet I have some difficulty to answer the second part of the question.Thank you

Cbarker11
 
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I'm pretty sure the square root function would succumb to http://mathhelpboards.com/math-notes-49/method-proving-some-non-linear-limits-4149.html of proving nonlinear limits.
 
Put $x=t^2,$ so $\displaystyle\lim_{x\to9}\sqrt x=\lim_{t\to3}t=3.$ Since the original limit is positive, we require that $|t|=t.$
So you can prove your result using the latter limit.
 

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