# \lim\limits_{x \to 5} \frac{\sqrt{3x+10}-5}{x^{2}-25}

## Question:

{eq}\lim\limits_{x \to 5} \frac{\sqrt{3x+10}-5}{x^{2}-25} {/eq}

Let's multiply the numerator and the denominator by the conjugate of the numerator, {eq}\sqrt{3x+10} +5: {/eq}

{eq}\begin{array}{l} \displaystyle \lim_{x \to 5} \dfrac{\sqrt{3x+10}-5}{x^{2}-25} \\ = \displaystyle \lim_{x \to 5} \dfrac{\sqrt{3x+10}-5}{x^{2}-25} \cdot \left( \dfrac{\sqrt{3x+10} +5}{\sqrt{3x+10} +5} \right) \\ = \displaystyle \lim_{x \to 5} \dfrac{(3x+10)-25}{(x^{2}-25)(\sqrt{3x+10} +5)} \\= \displaystyle \lim_{x \to 5} \dfrac{3x-15}{(x^{2}-25)(\sqrt{3x+10} +5)} \end{array} {/eq}

Note that it is still an indeterminate form of type 0/0. But we can simply this expression by factoring the numerator and the denominator:

{eq}\begin{array}{l} \displaystyle \lim_{x \to 5} \dfrac{3x-15}{(x^{2}-25)(\sqrt{3x+10} +5)} \\ = \displaystyle \lim_{x \to 5} \dfrac{3(x-5)}{(x-5)(x+5))(\sqrt{3x+10} +5)} \\ = \displaystyle \lim_{x \to 5} \dfrac{3}{(x+5)(\sqrt{3x+10} +5)} \end{array} {/eq}

Now we can apply direct substitution:

{eq}\displaystyle \lim_{x \to 5} \dfrac{3}{(x+5)(\sqrt{3x+10} +5)} = \dfrac { 3}{(5+5)(\sqrt{15+10}+ 5)} = \dfrac{3}{10\cdot(5+5)} = \dfrac 3 {100} {/eq}