Find the following limit:$$\lim\limits_{x\to \:4}\frac{\sqrt{x+5}-3}{x-4}$$
I tried to multiply by the conjugate and it did not work.
If you multiply by the conjugate, you should find
$$\frac{(x + 5) - 9}{(x - 4)(\sqrt{x + 5} + 3)} = \frac{x - 4}{(x - 4)\sqrt{x + 5} + 3}$$
Can you finish?
In case you are familiar with derivatives: this limit is the derivative of $f(x)=\sqrt{x+5}$ at $x=4$.
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If you multiply by the conjugate, you should find
$$\frac{(x + 5) - 9}{(x - 4)(\sqrt{x + 5} + 3)} = \frac{x - 4}{(x - 4)\sqrt{x + 5} + 3}$$
Can you finish?