Question

find the lim 1-cosx/x²
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Answer 1

Correct QuestioN :

Find the value of $\displaystyle \tt\lim_{x \to 0} \frac{1 - cos\,x}{x^2}$

AnsweR :

1 / 2.

SolutioN :

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{1 - cos\,x}{x^2}$

Here, By Trigonometry Formula.

$\tt \bullet \: \: \: \: \: Cos\,2\theta = 1 - 2Sin^2\,\theta$

$\tt \bullet \: \: \: \: \: Cos\,2\theta = 2 Cos^2\,\theta-1$

$\tt \bullet \: \: \: \: \: Cos\,2\theta = \frac{ 1-tan^2\,\theta }{1+tan^2\,\theta}$

According to question,

Let try to convert ( Cos 2θ ) in terms of 1 - cos θ.

$\tt \implies Cos\,2\theta = 1 - 2Sin^2\,\theta$

We can also write as,

$\tt \implies Cos\,\theta = 1 - 2Sin^2\, \frac{\theta}{2}$

$\tt \implies 1 - Cos\,\theta = 2Sin^2\, \frac{\theta}{2}$

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Now, putting the value of 1 - Cos x.

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{2Sin^2\, \frac{x}{2}}{x^2}$

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{2Sin^2\, \frac{x}{2}}{4 (\frac{x}{2}) ^2}$

Now, We know that,

$\dagger \: \: \: \: \: \: \displaystyle \tt\lim_{x \to 0} \frac{Sin\,x}{x} = 1.$

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{2 \bigg(Sin\, \frac{x}{2} \bigg)^2}{4 (\frac{x}{2}) ^2}$

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{2 \cancel{\bigg(Sin\, \frac{x}{2} \bigg)^2}}{4 \cancel{(\frac{x}{2}) ^2}}$

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{2 }{4}$

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{1 }{2}$

Therefore, the required value of given lim is 1 / 2.

Answer 2

✭ Appropriate Question ✭

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• ➠ find the lim 1-cosx/x² ?

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✪ Solution ✪

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• ➠ lim_(x->0) (1 - cos(x))/x^2

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$\large\underline{\red{\sf \orange{\bigstar} .}}$ Hint:-

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• ➠ Indeterminate form of type 0/0. Apply l'Hôpital's rule.

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☯ Applying l'Hôpital's rule, we get that

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• ➠ lim_(x->0) (1 - cos(x))/x^2 = lim_(x->0) d/( dx)(1 - cos(x))/( d/( dx) x^2) = lim_(x->0) sin(x)/(2 x) invisible comma

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➠ lim_(x->0) sin(x)/(2 x)

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$\large\underline{\red{\sf \red{\bigstar} .}}$ Hint:-

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• • ➠ Factor a constant multiple out of the limit.

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• • ➠ lim_(x->0) sin(x)/(2 x) = 1/2 (lim_(x->0) sin(x)/x):

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➠ 1/2 lim_(x->0) sin(x)/x

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$\large\underline{\red{\sf \green{\bigstar} .}}$ Hint:-

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• Indeterminate form of type 0/0. Apply l'Hôpital's rule.

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• ➠ lim_(x->0) sin(x)/x | = | lim_(x->0) ( d/( dx) sin(x))/(( dx)/( dx))

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• ➠ | = | lim_(x->0) cos(x)/1

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• ➠ | = | lim_(x->0) cos(x) invisible comma

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➠ 1/2lim_(x->0) cos(x)

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$\large\underline{\red{\sf \pink{\bigstar} .}}$ Hint:-

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• | The limit of a continuous function at a point is just its value there.

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• lim_(x->0) cos(x) = cos(0) = 1:

$\large\underline{\red{\sf \orange{\bigstar} .}}$ Answer: 1/2 $\large\underline{\red{\sf \orange{\bigstar} .}}$

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