30,48,68,90,.... What is the resoective series number after 90

Answer:\large\underline{\sf{Solution-}} Solution− Given that function\begin{gathered}\rm \: f(x) = \begin{cases} \dfrac{ \sqrt{(1 + Px) } - \sqrt{(1 - Px)}}{ x} \: \: \: ; - 1 \leq x < 0 \\ \\ \dfrac{2x + 1}{x - 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ;0 \leq x \leq 1\end{cases} \\ \end{gathered} f(x)= ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧ x(1+Px) − (1−Px) ;−1≤x<0x−22x+1 ;0≤x≤1 is continuous on [ - 1, 1 ].We know, A function f(x) is continuous at x = a, iff\begin{gathered}\color{green}\boxed{ \rm{ \:\rm \: \displaystyle\lim_{x \to a^-}\rm f(x) = \displaystyle\lim_{x \to a^ + }\rm f(x) = f(a) \: \: }} \\ \end{gathered} x→a − lim f(x)= x→a + lim f(x)=f(a) Since, it is given that function is continuous at x = 0.So,\begin{gathered}\rm \: f(0) = \displaystyle\lim_{x \to 0^-}\rm f(x) \\ \end{gathered} f(0)= x→0 − lim f(x) \begin{gathered}\rm \: \dfrac{2(0)+ 1}{0 - 2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{ \sqrt{(1 + Px) } - \sqrt{(1 - Px)}}{ x} \\ \end{gathered} 0−22(0)+1 = x→0 − lim x(1+Px) − (1−Px) \begin{gathered}\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{ \sqrt{(1 + Px) } - \sqrt{(1 - Px)}}{ x} \times \frac{\sqrt{(1 + Px)} + \sqrt{(1 - Px)}}{\sqrt{(1 + Px)} + \sqrt{(1 - Px)}} \\ \end{gathered} − 21 = x→0 − lim x(1+Px) − (1−Px) × (1+Px) + (1−Px) (1+Px) + (1−Px) \begin{gathered}\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{\bigg(\sqrt{(1 + Px)}\bigg)^{2} - \bigg(\sqrt{(1 - Px)}\bigg)^{2} }{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)} \\ \end{gathered} − 21 = x→0 − lim x( (1+Px) + (1−Px) )( (1+Px) ) 2 −( (1−Px) ) 2 \begin{gathered}\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{(1 + Px) - (1 - Px)}{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)} \\ \end{gathered} − 21 = x→0 − lim x( (1+Px) + (1−Px) )(1+Px)−(1−Px) \begin{gathered}\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{1 + Px - 1 + Px}{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)} \\ \end{gathered} − 21 = x→0 − lim x( (1+Px) + (1−Px) )1+Px−1+Px \begin{gathered}\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{2Px}{x\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)} \\ \end{gathered} − 21 = x→0 − lim x( (1+Px) + (1−Px) )2Px \begin{gathered}\rm \: - \dfrac{1}{2} = \displaystyle\lim_{x \to 0^-}\rm \dfrac{2P}{\bigg(\sqrt{(1 + Px)} + \sqrt{(1 - Px)}\bigg)} \\ \end{gathered} − 21 = x→0 − lim ( (1+Px) + (1−Px) )2P Now, to evaluate this limit, we use method of Substitution.So, Substitute\begin{gathered}\rm \: x \: = \: 0 - h \: = \: - \: h, \: \: as \: x \: \to \: 0 \: \: as \: \: h \: \to \: 0 \\ \end{gathered} x=0−h=−h,asx→0ash→0 So, above expression can be rewritten as\begin{gathered}\rm \: - \dfrac{1}{2} = \displaystyle\lim_{h \to 0}\rm \dfrac{2P}{\bigg(\sqrt{(1 - Ph)} + \sqrt{(1 + Ph)}\bigg)} \\ \end{gathered} − 21 = h→0lim ( (1−Ph) + (1+Ph) )2P \begin{gathered}\rm \: - \dfrac{1}{2} = \dfrac{2P}{ \sqrt{1 - 0} + \sqrt{1 + 0} } \\ \end{gathered} − 21 = 1−0 + 1+0 2P \begin{gathered}\rm \: - \dfrac{1}{2} = \dfrac{2P}{ 1 + 1} \\ \end{gathered} − 21 = 1+12P \begin{gathered}\rm \: - \dfrac{1}{2} = \dfrac{2P}{2} \\ \end{gathered} − 21 = 22P \begin{gathered}\color{green}\rm\implies \:\boxed{ \rm{ \:P \: = \: - \: \dfrac{1}{2} \: \: }} \\ \end{gathered} ⟹ P=− 21 \rule{190pt}{2pt}Additional Information :-\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered} MoreFormulaeMoreFormulae ★ x→0lim xsinx =1★ x→0lim xtanx =1★ x→0lim xlog(1+x) =1★ x→0lim xe x −1 =1★ x→0lim xa x −1 =loga