Ten limits: indeterminate forms and notable limits

Text

Evaluate the following limits (if they exist). $\text{a)}\ \lim_{x\to2}\frac{x^2-5x+6}{-2x^2+x+6}\qquad \text{b)}\ \lim_{x\to0}\frac{\sin(2x^2)}{\cos(3x)-1}$ $\text{c)}\ \lim_{x\to1}\frac{\ln(x^4)}{1-e^{1-x}}\qquad \text{d)}\ \lim_{x\to2}\frac{\sqrt{1-x^2}}{\sin(x-2)}$ $\text{e)}\ \lim_{x\to0}\frac{1-\sqrt[3]{1-2x}}{1-\sqrt{1+4x}}\qquad \text{f)}\ \lim_{x\to0^-}\sin\!\left(\frac1x\right)$ $\text{g)}\ \lim_{x\to+\infty}\sin x\qquad \text{h)}\ \lim_{x\to1^+}x^{\frac{2}{x-1}}$ $\text{i)}\ \lim_{x\to1^+}\frac{\tan(x-1)}{|1-x|}\qquad \text{j)}\ \lim_{x\to+\infty}\frac{\ln x-x^3+2}{2x^2-3x^3+\sin x}$

Solution

a) Factor: numerator $x^2-5x+6=(x-2)(x-3)$; denominator $-2x^2+x+6=-(2x^2-x-6)=-(2x+3)(x-2)$. Cancelling $(x-2)$: $\frac{(x-2)(x-3)}{-(2x+3)(x-2)}=\frac{x-3}{-(2x+3)}\xrightarrow{x\to2}\frac{-1}{-7}=\frac17.$

b) Using expansions: $\sin(2x^2)\sim 2x^2$ and $\cos(3x)-1\sim-\dfrac{(3x)^2}{2}=-\dfrac{9x^2}{2}$. Hence $\frac{2x^2}{-\frac{9}{2}x^2}=-\frac{4}{9}.$

c) $\ln(x^4)=4\ln x$. Set $t=x-1\to0$: $\ln x=\ln(1+t)\sim t$ and $1-e^{1-x}=1-e^{-t}\sim t$. Thus $\frac{4t}{t}\to 4.$

d) Near $x=2$ we have $1-x^2<0$ (at $x=2$ it equals $-3$): $\sqrt{1-x^2}$ is not defined in a neighbourhood of $2$. The limit does not exist in $\mathbb{R}$.

e) Expansions: $\sqrt[3]{1-2x}\sim 1-\dfrac{2x}{3}$, so the numerator $\sim\dfrac{2x}{3}$; $\sqrt{1+4x}\sim 1+2x$, so the denominator $\sim -2x$. Ratio $\frac{2x/3}{-2x}=-\frac13.$

f) $\sin(1/x)$ as $x\to0^-$ oscillates between $-1$ and $1$: the limit does not exist.

g) $\sin x$ as $x\to+\infty$ oscillates: the limit does not exist.

h) Form $1^{\infty}$. $\ln L=\lim_{x\to1^+}\dfrac{2}{x-1}\ln x$; with $t=x-1\to0^+$, $\ln x\sim t$, so $\dfrac{2}{t}\cdot t=2$. Therefore $L=e^2\approx7.389$.

i) As $x\to1^+$: $x-1\to0^+$, $|1-x|=x-1$ and $\tan(x-1)\sim x-1$. Hence $\frac{x-1}{x-1}\to 1.$

j) Dominant terms $\dfrac{-x^3}{-3x^3}=\dfrac13$ (logarithm and sine negligible). So the limit is $\dfrac13$.

$\boxed{a=\tfrac17,\ b=-\tfrac49,\ c=4,\ d=\nexists,\ e=-\tfrac13,\ f=\nexists,\ g=\nexists,\ h=e^2,\ i=1,\ j=\tfrac13}$