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Evaluate $\displaystyle\lim_{h\to 0}\frac{\sqrt{9+h}-3}{h}$.
Answer
\[\begin{align*} \lim_{h\to 0}\frac{\sqrt{9+h}-3}{h} &= \lim_{h\to 0}\frac{(\sqrt{9+h}-3)(\sqrt{9+h}+3)}{h(\sqrt{9+h}+3)}\text{ (rationalization)}\\\\ &= \lim_{h\to 0}\frac{9+h-9}{h(\sqrt{9+h}+3)} = \lim_{h\to 0}\frac{1}{\sqrt{9+h}+3}\\\\ &= 1. \end{align*}\]Evaluate $\displaystyle\lim_{h\to 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$.
Answer
\[\begin{align*} \lim_{h\to 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h} &= \lim_{h\to 0}\frac{x^2-(x+h)^2}{h(x+h)^2x^2} \\\\ &= \lim_{h\to 0}\frac{-2xh-h^2}{h(x+h)^2x^2} = \lim_{h\to 0}\frac{-2x-h}{(x+h)^2x^2} \\\\ &= \frac{-2}{x^3}. \end{align*}\]Prove that $\displaystyle\lim_{x\to 0^+}\sqrt{x}e^{\sin(\pi/x)}=0$.
Answer
Since $\sin(\pi/x)$ is bounded between $-1$ and $1$ for $x\neq 0$, we have \(e^{-1}\leq e^{\sin(\pi/x)}\leq e,\) and \(e^{-1}\sqrt{x}\leq \sqrt{x}e^{\sin(\pi/x)}\leq e\sqrt{x}.\) Since both $e^{-1}\sqrt{x}$ and $e\sqrt{x}$ converges to $0$, as $x$ approaches $0$ from the right, we can conclude that the desired limit is zero by the sqeeze theorem.
Use a graph to find a number $\delta$ such that
Answer
Referring to the graph, it becomes apparent that the function $f(x) = \sqrt{x^2+5}$ exceeds $2.7$ when $x$ is greater than $(2.7^2-5)^{1/2} = 1.5132745950421558…$, and it falls below $3.3$ when $x$ is less than $(3.3^2-5)^{1/2} = 2.4269322199023193…$. As a result, we can select $\delta = \min\left(2-(2.7^2-5)^{1/2}, (3.3^2-5)^{1/2}-2\right) = (3.3^2-5)^{1/2}-2$.
| Evaluate $\displaystyle\lim_{t\to 0}\left(\frac{1}{t\sqrt{1+ | t | }-\frac{1}{t}}\right)$. |
Answer
| $\displaystyle\lim_{t\to0}\frac{1}{t\sqrt{1+ | t | }-\frac{1}{t}}=\lim_{t\to 0}\frac{t}{t^2\sqrt{1+ | t | }-1}=\frac{0}{0\sqrt{1}-1}=0$ |