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This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.

The exercise is:

If $f'(a)=1$ for $a>0$, find $\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}$.

What came to my mind was to rationalize the denominator.

$$\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}$$

$$=\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}\cdot \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}$$

$$=\lim_{x \to a} \frac{(f(x)-f(a))(\sqrt{x}+\sqrt{a})}{x-a}$$

$$=\lim_{x \to a} \left(\frac{f(x)-f(a)}{x-a}\cdot (\sqrt{x}+\sqrt{a})\right)$$

$$=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=f'(a)\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=1\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=\lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=\sqrt{a}+\sqrt{a}$$

$$=2\sqrt{a}$$

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You made a mistake. (It's not a big one but it impacts your answer).

On Line $3$, you should have had $\sqrt{x}+\sqrt{a}$ in numerator.

Now, redo the steps (easy), and you'll end up with $2\sqrt{a}$.

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  • $\begingroup$ great.. i will fix it.. thanks so much! $\endgroup$ – gi2302 Sep 14 at 22:35

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