1 1 vote Evaluate the limit: \[\lim_{{x \to 0}} \frac{\ln \left(\left(x^2+1\right) \cos x\right)}{x^2}\] Calculus gate2024-da-memory-based goclasses calculus limits numerical-answers + – GO Classes 1.9k views answer comment Share Follow Print See 1 comment 1 1 comment reply SivaKaliKeshav commented Feb 5, 2024 reply Follow flag Some standard limits and property of logarithms, $1)\lim_{x \to 0} \dfrac{\tan x}{x} = 1$ $2)\lim_{x \to 0} \dfrac{\ln (x + 1) }{x} = 1$ $3) \ln (ab) = \ln a + \ln b$ Given limit , $\lim_{x \to 0} \dfrac{\ln ((x^2 +1) \cos x)}{x^2}$ we can rewrite it as, $\lim_{x \to 0} \left(\dfrac{\ln (x^2 +1)}{x^2} + \dfrac{\ln (\cos x)}{x^2}\right)$ now lets evaluate individually, Let $y=x^2$ ,as $x \to 0$ $y \to 0$, for the limit below we can use substitution with $y=x^2$ $\lim_{x \to 0}\dfrac{\ln (x^2 +1)}{x^2}$ $=\lim_{y \to 0}\dfrac{\ln (y +1)}{y}$ now the limit is in form of the first standard limit which is mentioned above so, $=\lim_{y \to 0}\dfrac{\ln (y +1)}{y}$$ =1$ second limit is, $\lim_{x \to 0}\dfrac{\ln (\cos x)}{x^2}$ $\left(\dfrac{0}{0} form \right)$, so apply L'Hopital rule, =$\lim_{x \to 0}-\dfrac{\tan x }{2x}$ $= -\dfrac{1}{2}\lim_{x \to 0}\dfrac{\tan x }{x}$ (now its in form of second standard limit mentioned above so,) $= -\dfrac{1}{2}\lim_{x \to 0}\dfrac{\tan x }{x}$$= -\dfrac{1}{2}(1)$ $= -\dfrac{1}{2}$ as both these limits exists we can use sum rule of limits as, $\lim_{x \to 0} \left(\dfrac{\ln (x^2 +1)}{x^2} + \dfrac{\ln (\cos x)}{x^2}\right)$ $=\lim_{x \to 0}\dfrac{\ln (x^2 +1)}{x^2} + \lim_{x \to 0}\dfrac{\ln (\cos x)}{x^2}$$= 1 - \dfrac{1}{2}$$= \dfrac{1}{2}$. 6 6 replyShare Please log in or register to add a comment.
1 1 vote Answer is 1/2 amityadav answered Feb 5, 2024 • edited Feb 5, 2024 by amityadav amityadav comment Share Follow See all 4 Comments 4 4 Comments reply pankaj kumar 70m commented Feb 5, 2024 i edited by pankaj kumar 70m Feb 7, 2024 reply Follow flag Answer : 1/2 0 0 replyShare SivaKaliKeshav commented Feb 5, 2024 reply Follow flag please read the question again, $\ln$ is for whole $(x^2+1)\cos x$ not just only for $(x^2+1)$. 1 1 replyShare pankaj kumar 70m commented Feb 6, 2024 reply Follow flag True , i got the mistake. 0 0 replyShare SivaKaliKeshav commented Feb 6, 2024 reply Follow flag and try to change your answer. 1 1 replyShare Please log in or register to add a comment.
0 0 votes ALWAYS REMEMBER WE DON'T HAVE TO JUST DIFFERENTIATE WE HAVE TO DIFFERENTIATE NUMERATOR / DENOMINATOR ALAG ALAG Sonu123x answered Feb 28, 2024 Sonu123x comment Share Follow 0 reply Please log in or register to add a comment.
