GATE DS&AI 2024 | Question: 50

Question

Evaluate the following limit:
\[
\lim _{x \rightarrow 0} \frac{\ln \left(\left(x^{2}+1\right) \cos x\right)}{x^{2}}=
\]

Answer 1

To find \( \lim_{{x \to 0}} \frac{\ln((x^2 + 1) \cdot \cos x)}{x^2} \), we can use L'Hôpital's Rule, which states that if \( \lim_{{x \to c}} f(x) = \lim_{{x \to c}} g(x) = 0 \) or \( \pm \infty \), and \( g'(x) \neq 0 \) near \( c \), then:

\[ \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \]

Let's find the derivatives of the numerator and denominator:

\[ f(x) = \ln((x^2 + 1) \cdot \cos x) \]

\[ f'(x) = \frac{1}{(x^2 + 1) \cdot \cos x} \cdot ((x^2 + 1)' \cdot \cos x + (x^2 + 1) \cdot (\cos x)') \]

\[ = \frac{1}{(x^2 + 1) \cdot \cos x} \cdot (2x \cdot \cos x - (x^2 + 1) \cdot \sin x) \]

\[ g(x) = x^2 \]

\[ g'(x) = 2x \]

Now, let's find \( \lim_{{x \to 0}} \frac{f'(x)}{g'(x)} \):

\[ \lim_{{x \to 0}} \frac{f'(x)}{g'(x)} = \lim_{{x \to 0}} \frac{\frac{1}{(x^2 + 1) \cdot \cos x} \cdot (2x \cdot \cos x - (x^2 + 1) \cdot \sin x)}{2x} \]

\[ = \lim_{{x \to 0}} \frac{2x \cdot \cos x - (x^2 + 1) \cdot \sin x}{2x \cdot (x^2 + 1) \cdot \cos x} \]

Now, we can apply L'Hôpital's Rule again:

\[ = \lim_{{x \to 0}} \frac{2 \cos x - 2x \cdot \sin x - 2x \cdot \sin x - (x^2 + 1) \cdot \cos x}{2 \cdot (x^2 + 1) \cdot \cos x - 4x^2 \cdot \sin x} \]

\[ = \lim_{{x \to 0}} \frac{2 \cos x - 4x \sin x - (x^2 + 1) \cos x}{2 \cos x - 4x^2 \sin x} \]

Now, plugging in \( x = 0 \) directly gives:

\[ = \frac{2 \cos 0 - 4 \cdot 0 \cdot \sin 0 - (0^2 + 1) \cdot \cos 0}{2 \cdot \cos 0 - 4 \cdot 0^2 \cdot \sin 0} \]

\[ = \frac{2 \cdot 1 - 0 - 1 \cdot 1}{2 \cdot 1 - 0} \]

\[ = \frac{2 - 1}{2} \]

\[ = \frac{1}{2} \]
So, \( \lim_{{x \to 0}} \frac{\ln((x^2 + 1) \cdot \cos x)}{x^2} = \frac{1}{2} \) or 0.5 .

Comments

you could save time by using $\log mn = \log m + \log n$ and the expansion of $\log(1+x).$

Answer 2

Ans : 1/2

We will apply L'Hopital Rule but little smartly.

ln((x² + 1) cos(x)) = ln(x²+1) + ln(cos(x)

distribute denominator x²  

ln(x²+1) / x² + ln(cos(x)/x² 

now apply L'Hopital seperatly on both the terms 

lim_x->0 2x / x²+1 . (2x)   +  lim_x->0  ^_sin(x) / cos(x) (2x)   

put the limit x=0   [  lim_x->0  Sin(X) / X = 1 ]

1 ^_ 1/2  = 1/2

 

 

Answer 3

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$

 

2) Second limit is,  $\lim_{x \to 0}\dfrac{\ln (\cos x)}{x^2}$  $\left(\dfrac{0}{0} \text{form} \right)$, 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}$.