Question

solve it fast........​

Answer 1

ʀᴇǫᴜɪʀᴇᴅ ᴀɴsᴡᴇʀ:–

$\begin{gathered} \qquad \quad{:}\longrightarrow \sf{\lim_{x \to 0} \bigg(\dfrac{1}{x^{2} + a}\bigg) + \lim_{x \to 0} \bigg(\dfrac{1}{a - x^{2}}\bigg)} \\ \\\end{gathered}$

• By applying the sum rule for limits in the equation, we get :

$\qquad \large { \underbrace { \bf{Quotient \: rule \: for \: limits: - }}}$

$\begin{gathered}\boxed{\sf{\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}}} \\ \\ \end{gathered}$

$\qquad \quad{:}\longrightarrow \sf{\dfrac{\lim_{x \to 0}(1)}{\lim_{x \to 0}(x^{2} + a)} + \dfrac{\lim_{x \to 0}(1)}{\lim_{x \to 0}(a - x^{2})}}$

• By applying the constant rule limits in the equation, we get :

$\qquad \large { \underbrace { \bf{Constant \: rule \: for \: limits: - }}}$

$\boxed{\sf{\lim_{x \to a} (c) = c}}$

$\qquad \quad{:}\longrightarrow \sf{\dfrac{1}{\lim_{x \to 0}(x^{2} + a)} + \dfrac{1}{\lim_{x \to 0}(a - x^{2})}}$

• By applying the sum rule for limits in the equation, we get :

$\qquad \large { \underbrace { \bf{Sum \: rule \: for \: limits: - }}}$

$\boxed{\sf{\lim_{x \to a} [f(x) + g(x)] =\lim_{x \to a} [f(x)] + \lim_{x \to a} [g(x)]}}$

$\qquad \quad:\longrightarrow \sf{\dfrac{1}{\lim_{x \to 0}(x^{2}) + \lim_{x \to 0}(a)} + \dfrac{1}{\lim_{x \to 0}(a) - \lim_{x \to 0}(x^{2})}}$

$\qquad \quad:\longrightarrow \sf{\dfrac{1}{0^{2} + a} + \dfrac{1}{a - 0^{2}}} \:\:\:\: [\because \sf{\lim_{x \to a} c = c}]$

$\qquad \quad:\longrightarrow\sf{\dfrac{1}{a} + \dfrac{1}{a}} \:\:\: \bigg[\because \sf{\dfrac{b}{a} + \dfrac{c}{a} = \dfrac{b + c}{a}\bigg]}$

$\qquad \quad:\longrightarrow\sf{\dfrac{(1 + 1)}{a} = \dfrac{2}{a}}$

$\begin{gathered}\boxed{\therefore \sf{\lim_{x \to 0} \bigg(\dfrac{1}{x^{2} + a}\bigg) + \lim_{x \to 0} \bigg(\dfrac{1}{a - x^{2}}\bigg) = \dfrac{2}{a}}} \\ \\ \end{gathered}$

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