$$\boxed{\displaystyle \lim_{x \to \infty} (\sqrt{ax^2+bx+c}-\sqrt{ax^2+px+q}) = \dfrac{b-p} {2\sqrt{a}} }$$In this case, it is given that The value of $\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{18x^2-x+1}-3x} {\sqrt{x^2+2x}}$ is $\cdots \cdot$ You should also note Find the value of $\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{8x^2+1}} {x^2+4}$. $\displaystyle \lim_{x \to \infty} \sqrt{\left(\dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x} + 2\sqrt{2}}\right)^2}$ (Answer C), Problem Number 34  Why is the zh (ʒ) sound so infrequent in English? c. $\displaystyle \lim_{x \to \infty} \dfrac{\csc \frac{1}{x}} {x} $, Answer a)  $\displaystyle \lim_{x \to \infty} \dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}- 2\sqrt{x} + 2\sqrt{2}}$ c. $\displaystyle \lim_{x \to \infty} (\sqrt{x+5}-\sqrt{2x-3})$, You should recall that The coefficients of leading terms in numerator and denominator are $3$ and $-5$, respectively. $\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{\sin \dfrac{3}{x}} {\left(1-\cos \dfrac{2}{x} \right) \cdot x^2 \cdot \sin \dfrac{1}{x}} \\ & = \lim_{y \to 0} \dfrac{\sin 3y} {(1-\cos 2y) \cdot \left(\dfrac{1}{y}\right)^2 \cdot \sin y} \\ & = \lim_{y \to 0} \dfrac{\sin 3y \cdot y^2}{(1-(1-2 \sin^2 y)) \cdot \sin y} \\ & = \dfrac12 \lim_{y \to 0} \dfrac{\sin 3y} {\sin y} \cdot \dfrac{y}{\sin y} \cdot \dfrac{y}{\sin y} \\ & = \dfrac12 \times 3 \times 1 \times 1 = \dfrac32 \end{aligned}$ Here are the “all-out” problems on limit at infinity. B. B. a. Divide each term by $3^{2n}$, then apply the limit at infinity theorems. But if f(t) becomes arbitrarily large only $C_1=0$ satisfies this condition and you get the trivial solution, ODE with limit boundary condition at infinity, Stack Overflow for Teams is now free for up to 50 users, forever, Mysterious inconsistency in an inhomogeneous linear 2nd order ODE with specified boundary values, Transforming an ODE with final condition to an ODE with an initial condition, Solving second order nonlinear ODE given boundary condition at infinity. Recall that means becomes arbitrarily close to as long as is sufficiently close to We can extend this idea to limits at infinity. Here are the “all-out” problems on limit at infinity. Answer a)  Because $x$ tends to infinity, the expression $\left(x+\dfrac{1}{7}\right)^7$ will getting bigger in value at a much faster rate, causing $O(x^5)$ can be ignored. (Answer E), Problem Number 15 I think i didn't phrase my question properly, the purpose is to use the boundary condition in finding the solution at time (t). $\begin{aligned} \displaystyle \lim_{x \to \infty}-(3x^2 + 9) & =-(3(\infty)^2 + 9) \\ & =-(\infty + 9) =-\infty \end{aligned}$, Problem Number 2 $\displaystyle \lim_{x \to \infty} (-x + 4) =-\infty + 4 =-\infty$ Solutions of problems with boundary conditions involving a limit at infinity can be difficult. $\displaystyle \lim_{x \to \infty} \sqrt{\dfrac{x^3 + \cdots}{2x^3 + \cdots}}$ Note: The denominator in the last term tends to zero, making the limits tends to infinity, but we don’t suggest you to simply write $\dfrac{0+1}{0-0} = \infty$, as the answer is not “infinity”, but “undefined” since the denominator is exactly zero. $-\dfrac{10}{3}$                   E. $-\dfrac{5}{3}\sqrt{2}$, By applying rationalizing method, we have, $$\begin{aligned} & \displaystyle \lim_{x \to \infty} 2x\left(\sqrt{9+\dfrac{10}{x}}- 3\right) \\ & = \lim_{x \to \infty} 2x\left(\sqrt{9+\dfrac{10}{x}}-3\right) \times \dfrac{\sqrt{9+\dfrac{10}{x}} + 3}{\sqrt{9+\dfrac{10}{x}} + 3} \\ & = \lim_{x \to \infty} \dfrac{2x \cdot \left(9+ \dfrac{10}{x}-3^2\right)}{\sqrt{9+\dfrac{10}{x}} + 3} \\ & = \lim_{x \to \infty} \dfrac{2 \cdot 10}{\sqrt{9+\dfrac{10}{x}} + 3} \\ & = \dfrac{20}{\sqrt{9+0} + 3} = \dfrac{10}{3} \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} 2x\left(\sqrt{9+\dfrac{10}{x}}-3\right) = \dfrac{10}{3}}$, The value of $\displaystyle \lim_{x \to \infty} \dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x} + 2\sqrt{2}}$ $= \cdots \cdot$, $\displaystyle \lim_{x \to \infty} \dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}- 2\sqrt{x} + 2\sqrt{2}}$, $\displaystyle \lim_{x \to \infty} \sqrt{\left(\dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x} + 2\sqrt{2}}\right)^2}$, $\displaystyle \lim_{x \to \infty} \sqrt{\dfrac{(x^2-4x + 4)(x + 2 + \sqrt{4x})}{(\sqrt{2}x^{\frac32}-2x^{\frac12} + 2\sqrt{2})^2}}$.
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