Question

Q1.evaluate the above limit​ please

Answer 1

Step-by-step explanation:

Solution :

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

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

⠀⠀⠀⠀⠀Quotient rule for limits :

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

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

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

⠀⠀⠀⠀⠀Constant rule for limits :

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

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

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

⠀⠀⠀⠀⠀Sum rule for limits :

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

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

$\begin{gathered}:\implies \sf{\dfrac{1}{0^{2} + a} + \dfrac{1}{a - 0^{2}}} \:\:\:\: [\because \sf{\lim_{x \to a} c = c}] \\ \\\end{gathered}$

$\begin{gathered}:\implies \sf{\dfrac{1}{a} + \dfrac{1}{a}} \:\:\: \bigg[\because \sf{\dfrac{b}{a} + \dfrac{c}{a} = \dfrac{b + c}{a}\bigg]}\\ \\\end{gathered}$

$</p><p>\begin{gathered}:\implies \sf{\dfrac{(1 + 1)}{a} = \dfrac{2}{a}} \\ \\\end{gathered}$

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