Question

Evaluate the following:
$1)\displaystyle \rm \lim_{x \to 0} \frac{1-cos x}{x^2}$
$2)\displaystyle \rm \lim_{x \to 0} \frac{ln(tanx+1)}{sinx}$

Answer 1

To Evaluate :

$\displaystyle \rm 1) \lim_{x \to 0} \frac{1-cosx}{x^2} \\2) \lim_{x \to 0}\frac{ln(tanx+1)}{sinx}$

Solution :

1) $\displaystyle \rm \lim_{x \to 0} \frac{1-cosx}{x^2}$

so, here if we put the value of x directly as zero, we will get the result as 0/0 which is an indeterminant form.

using the formula :

$\bull \rm \displaystyle \lim_{x\to 0} \frac{sinx}{x} = 1$

$\bull \rm \displaystyle \lim_{x \to 0} cosx =1$

So, we will first simplify this limit :

$\displaystyle \rm \implies \lim_{x \to 0} \frac{(1-cosx)(1+cosx)}{x^2(1+cosx)}$

$\displaystyle \rm \implies \lim_{x \to 0} \frac{1-cos^2x}{x^2(1+cosx)}$

$\implies\displaystyle \rm \lim_{x \to 0} \frac{sin^2x}{x^2(1+cosx)}$

$\implies \displaystyle \rm \lim_{x \to 0} \frac{sin^2}{x^2}\times \frac{1}{1+cosx}$

$\implies \displaystyle \rm \lim_{x \to 0} (\frac{sinx}{x} {)}^2\times\frac{1}{1+cosx}$

now, as x approaches 0, we get the result as:

$\implies \displaystyle \rm 1\times\frac{1}{1+1}$

$\displaystyle \rm \red{\underline{\boxed{ \implies \displaystyle \rm \lim_{x \to 0} \frac{1-cosx}{x^2} = \frac{1}{2}}}}$

Identities used here :

$\displaystyle \rm \bull (a+b)(a-b) = a^2-b^2$

$\displaystyle \rm \bull 1-cos^2\theta = sin^2 \theta$

$\displaystyle \rm \bull (\frac{x^2}{y^2} ) =(\frac{x}{y} )^{2}$

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2) $\displaystyle \rm \lim_{x \to 0}\frac{ln(tanx+1)}{sinx}$

using the formula :

$\bull \rm \displaystyle \rm \lim_{x \to 0} \frac{ln(x+1)}{x} = 1$

$\bull \rm \displaystyle \lim_{x \to 0} cosx =1$

now, simplifying this limit :

$\implies \rm \displaystyle \rm \lim_{x \to 0}\frac{ln(tanx+1)}{sinx\times \frac{tanx}{tanx} }$

$\implies \displaystyle \rm \lim_{x \to 0}\frac{ln(tanx+1)}{tanx}\times \frac{1}{cosx}$

$\implies \rm \displaystyle 1\times 1$

$\blue{\underline{\boxed{\implies \rm \displaystyle \rm \lim_{x \to 0}\frac{ln(tanx+1)}{sinx} = 1}}}$

Identities used here :

$\bull \displaystyle \rm \frac{sin\theta}{tan\theta} = cos\theta$

Answer 2

Given :  $\displaystyle \rm \lim_{x \to 0} \frac{1-\cos x}{x^2}$

$\displaystyle \rm \lim_{x \to 0} \frac{\ln (\tan x + 1)}{\sin x}$

To Find : evaluate the Limit

Solution:

$\displaystyle \rm \lim_{x \to 0} \frac{1-\cos x}{x^2}$

cos ( 0) = 1  => 1 - cos(0) = 0

 and 0² = 0

0/0  form

Hence Apply L'Hosptital Rule

Differentiate numerator and Denominator

$\displaystyle \rm \lim_{x \to 0} \frac{ \sin x}{2x}$

Still 0/0  form

Differentiate numerator and Denominator  again

$\displaystyle \rm \lim_{x \to 0} \frac{ \cos x}{2}$

= 1/2

 $\displaystyle \rm \lim_{x \to 0} \frac{1-\cos x}{x^2}$  = 1/2

$\displaystyle \rm \lim_{x \to 0} \frac{\ln (\tan x + 1)}{\sin x}$

tan 0 = 0  , sin 0 = 0  ln 1  = 0

0/0  form

On Differentiating numerator and Denominator

$\displaystyle \rm \lim_{x \to 0} \frac{ \sec^2 x }{(\tan x + 1)\cos x}$

sec 0 = 1

cos 0 = 1

= 1²/(0+ 1)1

= 1/1

= 1

$\displaystyle \rm \lim_{x \to 0} \frac{\ln (\tan x + 1)}{\sin x}$ = 1

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