Question

2+23 jsbsmBsbsbnnnsns. Ssjs

Answer 1

Answer:

25

Step-by-step explanation:

As far as I can understand your question is that only

2+23=25 only

Mark it's as brainliest

Answer 2

Step-by-step explanation:

GIVEN :–

• A limit –

$\begin{gathered} \\ \implies \bf \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) - \tan(3x)}{x}\bigg\}\\ \end{gathered}⟹x→0lim{xsin(2x)−tan(3x)}</p><p>TO FIND :–</p><p>• Value of limit = ?</p><p>SOLUTION :–</p><p>• Let –</p><p>\begin{gathered} \\ \implies \bf P = \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) - \tan(3x)}{x}\bigg\}\\ \end{gathered}⟹P=x→0lim{xsin(2x)−tan(3x)}</p><p>• Put the limit –</p><p>\begin{gathered} \\ \implies \bf P =\dfrac{ \sin(2 \times 0) - \tan(3 \times 0)}{0}\\ \end{gathered}⟹P=0sin(2×0)−tan(3×0)</p><p>\begin{gathered} \\ \implies \bf P =\dfrac{ \sin(0) - \tan(0)}{0}\\ \end{gathered}⟹P=0sin(0)−tan(0)</p><p>\begin{gathered} \\ \implies \bf P =\dfrac{0-0}{0}\\ \end{gathered}⟹P=00−0</p><p>\begin{gathered} \\ \implies \bf P =\dfrac{0}{0}\\ \end{gathered}⟹P=00</p><p>• It's undefined form. Now Let's solve –</p><p>\begin{gathered} \\ \implies \bf P = \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x)}{x}\bigg\} -\lim_{x \to0} \bigg\{ \dfrac{ \tan(3x)}{x}\bigg\}\\ \end{gathered}⟹P=x→0lim{xsin(2x)}−x→0lim{xtan(3x)}</p><p>• We can write this as –</p><p>\begin{gathered} \\ \implies \bf P =\lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) \times2}{(2x)}\bigg\} -\lim_{x \to0} \bigg\{ \dfrac{ \tan(3x) \times 3}{3x}\bigg\}\\ \end{gathered}⟹P=x→0lim{(2x)sin(2x)×2}−x→0lim{3xtan(3x)×3}</p><p>\begin{gathered} \\ \implies \bf P =2\lim_{x \to0} \bigg\{ \dfrac{ \sin(2x)}{(2x)}\bigg\} -3\lim_{x \to0} \bigg\{ \dfrac{ \tan(3x)}{3x}\bigg\}\\ \end{gathered}⟹P=2x→0lim{(2x)sin(2x)}−3x→0lim{3xtan(3x)}</p><p>• We also know that –</p><p>\begin{gathered} \\ \implies \large\bf {\red{ \lim_{x \to0} \bigg\{ \dfrac{ \sin x}{x}\bigg\} = 1 \: \: , \: \: \lim_{x \to0} \bigg\{ \dfrac{ \tan x}{x}\bigg\} = 1}}\\ \end{gathered}⟹x→0lim{xsinx}=1,x→0lim{xtanx}=1</p><p>• So that –</p><p>\begin{gathered} \\ \implies \bf P =2(1) -3(1)\\ \end{gathered}⟹P=2(1)−3(1)</p><p>\begin{gathered} \\ \implies \bf P =2-3\\ \end{gathered}⟹P=2−3</p><p>\begin{gathered} \\ \implies \large{ \boxed{\bf P =-1}}\\ \end{gathered}⟹P=−1</p><p>• Hence –</p><p>\begin{gathered} \\ \implies \large \pink{ \boxed{\bf \lim_{x \to0} \bigg\{ \dfrac{ \sin(2x) - \tan(3x)}{x}\bigg\} =-1}}\\ \end{gathered}$⟹x→0lim{xsin(2x)−tan(3x)}=−1