find the limit. (hint: try multiplying and dividing by the conjugate.)\nlim (√(x + 4) - √(x + 9))\nx→∞\nlim…

Explanation:

Step1: Multiply and divide by conjugate

Multiply and divide $\lim_{x\rightarrow\infty}(\sqrt{x + 4}-\sqrt{x + 9})$ by its conjugate $\sqrt{x + 4}+\sqrt{x + 9}$. [ \begin{align*} &\lim_{x\rightarrow\infty}(\sqrt{x + 4}-\sqrt{x + 9})\times\frac{\sqrt{x + 4}+\sqrt{x + 9}}{\sqrt{x + 4}+\sqrt{x + 9}}\ =&\lim_{x\rightarrow\infty}\frac{(\sqrt{x + 4})^2-(\sqrt{x + 9})^2}{\sqrt{x + 4}+\sqrt{x + 9}}\ =&\lim_{x\rightarrow\infty}\frac{(x + 4)-(x + 9)}{\sqrt{x + 4}+\sqrt{x + 9}} \end{align*} ]

Step2: Simplify the numerator

Simplify the numerator $(x + 4)-(x + 9)$: [ \begin{align*} (x + 4)-(x + 9)&=x+4 - x-9\ &=- 5 \end{align*} ] So the limit becomes $\lim_{x\rightarrow\infty}\frac{-5}{\sqrt{x + 4}+\sqrt{x + 9}}$.

Step3: Analyze the limit as $x\rightarrow\infty$

As $x\rightarrow\infty$, both $\sqrt{x + 4}\rightarrow\infty$ and $\sqrt{x + 9}\rightarrow\infty$. Then $\sqrt{x + 4}+\sqrt{x + 9}\rightarrow\infty$. We know that $\lim_{x\rightarrow\infty}\frac{-5}{\sqrt{x + 4}+\sqrt{x + 9}} = 0$ since the numerator is a non - zero constant and the denominator approaches infinity.

Answer:

$0$