Sketching Calc

[frosty tide]

Does anyone know how to solve this?

[keen zodiac]

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[open lake]

Let y = sqrt(x), factorize y^2 - a and then rewrite it in terms of x

[lunar coyote]

For part (b), we have a right-hand limit thus our original assumption, instead of $0<\abs{x-a}<\delta\implies 0<\abs{f(x)-L}<\varepsilon$, will be
$$0<x-a<\delta\implies\abs{f(x)-L}<\varepsilon$$
The rest follows like any $\varepsilon$-$\delta$ proof would be.

[lusty iron]

waffleqz

[lunar coyote]

That is for $\lim_{x\to a^+}f(x)=L$.

[lusty iron]

waffleqz

[young locust]

$$ \frac{|x-a|}{\sqrt{x}+\sqrt{a}} $$

[lusty iron]

aL

[young locust]

and bound the denominator around x=a