MHB Using the epsilon and delta definition to prove limit

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The limit of the square root function as x approaches 9 is determined to be 3. To prove this using the epsilon-delta definition, one can substitute x with t^2, simplifying the limit to lim(t→3)t=3. This approach confirms that the limit is valid since the original limit is positive, necessitating |t|=t. The discussion emphasizes the application of the epsilon-delta definition in proving limits for nonlinear functions. Understanding these concepts is crucial for mastering limit proofs in calculus.
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.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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