Limit using Binomial theorem Questions in English

(A) Using the binomial expansion for $(1+x)^n$ where $x \to 0$:
$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + ...$
Substituting this into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{(1 + nx + \frac{n(n-1)}{2}x^2 + ...) - 1}{x}$
$= \mathop {\lim }\limits_{x \to 0} \frac{nx + \frac{n(n-1)}{2}x^2 + ...}{x}$
$= \mathop {\lim }\limits_{x \to 0} (n + \frac{n(n-1)}{2}x + ...)$
$= n$
Alternatively,applying $L-Hospital's$ rule by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}((1+x)^n - 1)}{\frac{d}{dx}(x)} = \mathop {\lim }\limits_{x \to 0} \frac{n(1+x)^{n-1}}{1} = n(1+0)^{n-1} = n$.

(B) $\text{Given limit } L = \mathop {\lim }\limits_{n \to \infty } \sin (\pi \sqrt {{n^2} + 1} )$
$\text{Using the binomial expansion } (1+x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \dots \text{ for } |x| < 1:$
$L = \mathop {\lim }\limits_{n \to \infty } \sin \left( \pi n \sqrt {1 + \frac{1}{n^2}} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \left( \pi n \left( 1 + \frac{1}{2n^2} - \frac{1}{8n^4} + \dots \right) \right)$
$L = \mathop {\lim }\limits_{n \to \infty } \sin \left( n\pi + \frac{\pi}{2n} - \frac{\pi}{8n^3} + \dots \right)$
$\text{Since } \sin(n\pi + \theta) = (-1)^n \sin(\theta), \text{ we have:}$
$L = \mathop {\lim }\limits_{n \to \infty } (-1)^n \sin \left( \frac{\pi}{2n} - \frac{\pi}{8n^3} + \dots \right)$
$\text{As } n \to \infty, \text{ the argument of sine approaches } 0, \text{ so } \sin(0) = 0.$
$L = 0.$

(A) Consider the limit $L = \mathop {Lim}\limits_{x \to - \infty } [(x^5 + 7x^4 + 2)^c - x]$.
For the limit to be finite and non-zero, we factor out $x^{5c}$:
$L = \mathop {Lim}\limits_{x \to - \infty } [x^{5c}(1 + \frac{7}{x} + \frac{2}{x^5})^c - x]$.
For this to be of the form $\infty \times 0$, we must have $5c = 1$, so $c = 1/5$.
Substituting $c = 1/5$:
$L = \mathop {Lim}\limits_{x \to - \infty } x[(1 + \frac{7}{x} + \frac{2}{x^5})^{1/5} - 1]$.
Using the binomial expansion $(1+u)^n \approx 1 + nu$ for small $u$:
$L = \mathop {Lim}\limits_{x \to - \infty } x[1 + \frac{1}{5}(\frac{7}{x} + \frac{2}{x^5}) - 1]$.
$L = \mathop {Lim}\limits_{x \to - \infty } x[\frac{7}{5x} + \frac{2}{5x^5}] = \frac{7}{5}$.
Thus, $c = 1/5$ and the limit is $7/5$.

(A) Let $L = \mathop {\lim }\limits_{x \to \infty } {x^{\frac{1}{3}}}\left( {{{\left( {x + 1} \right)}^{\frac{2}{3}}} - {{\left( {x - 1} \right)}^{\frac{2}{3}}}} \right)$.
We can rewrite the expression as $L = \mathop {\lim }\limits_{x \to \infty } x \left( {\left( {1 + \frac{1}{x}} \right)^{\frac{2}{3}} - \left( {1 - \frac{1}{x}} \right)^{\frac{2}{3}}} \right)$.
Using the binomial expansion $(1+u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2$ for small $u$,where $u = \pm \frac{1}{x}$ and $n = \frac{2}{3}$:
$\left( {1 + \frac{1}{x}} \right)^{\frac{2}{3}} \approx 1 + \frac{2}{3x} + \frac{\frac{2}{3}(\frac{2}{3}-1)}{2x^2} = 1 + \frac{2}{3x} - \frac{1}{9x^2}$.
$\left( {1 - \frac{1}{x}} \right)^{\frac{2}{3}} \approx 1 - \frac{2}{3x} + \frac{\frac{2}{3}(\frac{2}{3}-1)}{2x^2} = 1 - \frac{2}{3x} - \frac{1}{9x^2}$.
Substituting these into the expression:
$L = \mathop {\lim }\limits_{x \to \infty } x \left( (1 + \frac{2}{3x} - \frac{1}{9x^2}) - (1 - \frac{2}{3x} - \frac{1}{9x^2}) \right)$.
$L = \mathop {\lim }\limits_{x \to \infty } x \left( \frac{4}{3x} \right) = \frac{4}{3}$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{x}{(1-\sin x)^{1/8}-(1+\sin x)^{1/8}}$.
Using the binomial expansion $(1+u)^n \approx 1+nu$ for small $u$,where $n = 1/8$:
$(1-\sin x)^{1/8} \approx 1 - \frac{1}{8} \sin x$.
$(1+\sin x)^{1/8} \approx 1 + \frac{1}{8} \sin x$.
Substituting these into the limit expression:
$L = \lim _{x \rightarrow 0} \frac{x}{(1 - \frac{1}{8} \sin x) - (1 + \frac{1}{8} \sin x)}$.
$L = \lim _{x \rightarrow 0} \frac{x}{-\frac{2}{8} \sin x} = \lim _{x \rightarrow 0} \frac{x}{-\frac{1}{4} \sin x}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have $\lim _{x \rightarrow 0} \frac{x}{\sin x} = 1$.
Therefore,$L = -4 \times 1 = -4$.

(B) We have $L = \lim_{x \rightarrow 0} \frac{a - a(1 - \frac{x^2}{a^2})^{1/2} - \frac{x^2}{4}}{x^4}$.
Using the binomial expansion $(1 - u)^{1/2} = 1 - \frac{1}{2}u - \frac{1}{8}u^2 - \dots$ where $u = \frac{x^2}{a^2}$:
$L = \lim_{x}$ ${\rightarrow 0} \frac{a - a(1 - \frac{1}{2}(\frac{x^2}{a^2}) - \frac{1}{8}(\frac{x^2}{a^2})^2) - \frac{x^2}{4}}{x^4}$
$L = \lim_{x \rightarrow 0} \frac{a - a + \frac{x^2}{2a} + \frac{x^4}{8a^3} - \frac{x^2}{4}}{x^4}$
$L = \lim_{x \rightarrow 0} \frac{x^2(\frac{1}{2a} - \frac{1}{4}) + \frac{x^4}{8a^3}}{x^4}$.
For the limit to be finite,the coefficient of $x^2$ must be zero:
$\frac{1}{2a} - \frac{1}{4} = 0 \implies a = 2$.
Substituting $a = 2$ into the expression:
$L = \lim_{x \rightarrow 0} \frac{\frac{x^4}{8(2)^3}}{x^4} = \frac{1}{8 \times 8} = \frac{1}{64}$.
Thus,$a = 2$ and $L = \frac{1}{64}$.
Therefore,options $(A)$ and $(C)$ are correct.

(A) To evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4}$,we use rationalization.
Multiply the numerator and denominator by the conjugate $\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}$:
$L = \lim _{x \rightarrow 0} \frac{(1+\sqrt{1+x^4})-(2+x^5+x^6)}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6})}$
$L = \lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1-x^5-x^6}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6})}$
Using the binomial expansion $\sqrt{1+u} \approx 1 + \frac{u}{2}$ for small $u$,where $u = x^4$,we have $\sqrt{1+x^4} \approx 1 + \frac{x^4}{2}$.
Substituting this into the limit:
$L = \lim _{x \rightarrow 0} \frac{1 + \frac{x^4}{2} - 1 - x^5 - x^6}{x^4(\sqrt{1+1} + \sqrt{2})} = \lim _{x \rightarrow 0} \frac{\frac{x^4}{2} - x^5 - x^6}{x^4(2\sqrt{2})}$
$L = \lim _{x \rightarrow 0} \frac{\frac{1}{2} - x - x^2}{2\sqrt{2}} = \frac{1/2}{2\sqrt{2}} = \frac{1}{4\sqrt{2}}$

(A) We want to evaluate $L = \lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)$.
First,factor out $x$ from each term:
$L = \lim _{x \rightarrow \infty} \left( x(1+\frac{4}{x})^{1/3} - x(1-\frac{3}{x})^{1/2} \right)$.
Using the binomial expansion $(1+u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2$ for small $u$:
$(1+\frac{4}{x})^{1/3} \approx 1 + \frac{1}{3}(\frac{4}{x}) = 1 + \frac{4}{3x}$.
$(1-\frac{3}{x})^{1/2} \approx 1 + \frac{1}{2}(-\frac{3}{x}) = 1 - \frac{3}{2x}$.
Substituting these into the limit:
$L = \lim _{x \rightarrow \infty} \left( x(1 + \frac{4}{3x}) - x(1 - \frac{3}{2x}) \right)$.
$L = \lim _{x \rightarrow \infty} \left( x + \frac{4}{3} - x + \frac{3}{2} \right)$.
$L = \frac{4}{3} + \frac{3}{2} = \frac{8+9}{6} = \frac{17}{6}$.

(B) Let $L = \lim _{x}$ ${\rightarrow \infty}\left\{x-\left(\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right)^{1/n}\right\}$.
Factor out $x$ from the expression inside the $n$-th root:
$L = \lim _{x}$ ${\rightarrow \infty} \left\{ x - x \left( \left(1-\frac{a_1}{x}\right)\left(1-\frac{a_2}{x}\right) \ldots\left(1-\frac{a_n}{x}\right) \right)^{1/n} \right\}$.
Using the binomial approximation $(1-u)^k \approx 1-ku$ for small $u$:
$L = \lim _{x}$ ${\rightarrow \infty} x \left\{ 1 - \left(1-\frac{a_1}{nx}\right)\left(1-\frac{a_2}{nx}\right) \ldots\left(1-\frac{a_n}{nx}\right) \right\}$.
Expanding the product and keeping terms up to $O(1/x)$:
$L = \lim _{x}$ ${\rightarrow \infty} x \left\{ 1 - \left( 1 - \frac{a_1+a_2+\ldots+a_n}{nx} + O\left(\frac{1}{x^2}\right) \right) \right\}$.
$L = \lim _{x \rightarrow \infty} x \left( \frac{a_1+a_2+\ldots+a_n}{nx} \right) = \frac{a_1+a_2+\ldots+a_n}{n}$.