Six limits: polynomials, notable limits, roots

Text

Evaluate the following limits. $\text{a)}\ \lim_{x\to0^+}\frac{3x-2x^2}{x-2x^3}\qquad \text{b)}\ \lim_{x\to+\infty}\frac{3x-2x^2}{x-2x^3}$ $\text{c)}\ \lim_{x\to0^+}\frac{\sin(x^2)}{1-\cos(2x)}\qquad \text{d)}\ \lim_{x\to0^+}x\ln(x^4)$ $\text{e)}\ \lim_{x\to0^+}\frac{1-\sqrt{1+2x}}{1-\sqrt{1+x}}\qquad \text{f)}\ \lim_{x\to+\infty}\frac{1-\sqrt{1+2x}}{1-\sqrt{1+x}}$

Solution

a) Factor out $x$: $\dfrac{x(3-2x)}{x(1-2x^2)}=\dfrac{3-2x}{1-2x^2}\xrightarrow{x\to0^+}\dfrac{3}{1}=3.$

b) Dominant terms $\dfrac{-2x^2}{-2x^3}=\dfrac{1}{x}\to 0$ as $x\to+\infty$.

c) Expansions: $\sin(x^2)\sim x^2$ and $1-\cos(2x)\sim\dfrac{(2x)^2}{2}=2x^2$. Hence $\dfrac{x^2}{2x^2}=\dfrac12.$

d) $x\ln(x^4)=4\,x\ln x$. Since $x\ln x\to 0$ as $x\to0^+$, the limit is $0$.

e) Expansions: $\sqrt{1+2x}\sim 1+x$, numerator $\sim -x$; $\sqrt{1+x}\sim 1+\dfrac{x}{2}$, denominator $\sim -\dfrac{x}{2}$. Ratio $\dfrac{-x}{-x/2}=2.$

f) As $x\to+\infty$: $\sqrt{1+2x}\sim\sqrt{2x}=\sqrt2\,\sqrt x$ and $\sqrt{1+x}\sim\sqrt x$. Ratio $\dfrac{-\sqrt2\,\sqrt x}{-\sqrt x}=\sqrt2.$

$\boxed{a=3,\ b=0,\ c=\tfrac12,\ d=0,\ e=2,\ f=\sqrt2\approx1.414}$