Question

find limit of
lim cosx/x+1
x=-1​

Answer 1

Answer:

limx→0cos(x)−1x=0 . We determine this by utilising L'hospital's Rule. To paraphrase, L'Hospital's rule states that

Answer 2

$Find the value of \displaystyle \tt\lim_{x \to 0} \frac{1 - cos\,x}{x^2} </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>1−cosx</p><p> </p><p>$

AnsweR:-

1 / 2.

Solution :-

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{1 - cos\,x}{x^2}:⟹ </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>1−cosx</p><p>$

Here, By Trigonometry Formula.

$\tt \bullet \: \: \: \: \: Cos\,2\theta = 1 - 2Sin^2\,\theta∙Cos2θ=1−2Sin </p><p>2</p><p> θ</p><p></p><p>\tt \bullet \: \: \: \: \: Cos\,2\theta = 2 Cos^2\,\theta-1∙Cos2θ=2Cos </p><p>2</p><p> θ−1</p><p></p><p>\begin{gathered}\tt \bullet \: \: \: \: \: Cos\,2\theta = \frac{ 1-tan^2\,\theta }{1+tan^2\,\theta} \\ \end{gathered} </p><p>∙Cos2θ= </p><p>1+tan </p><p>2</p><p> θ</p><p>1−tan </p><p>2</p><p> θ$

According to question

$Let try to convert ( Cos 2θ ) in terms of 1 - cos θ.</p><p></p><p>\tt \implies Cos\,2\theta = 1 - 2Sin^2\,\theta⟹Cos2θ=1−2Sin </p><p>2</p><p> θ</p><p></p><p>We can also write as,</p><p></p><p>\begin{gathered}\tt \implies Cos\,\theta = 1 - 2Sin^2\, \frac{\theta}{2} \\ \end{gathered} </p><p>⟹Cosθ=1−2Sin </p><p>2</p><p> </p><p>2</p><p>θ</p><p>$

$\begin{gathered}\tt \implies 1 - Cos\,\theta = 2Sin^2\, \frac{\theta}{2} \\ \end{gathered} </p><p>⟹1−Cosθ=2Sin </p><p>2</p><p> </p><p>2</p><p>θ</p><p> </p><p>$

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Now, putting the value of 1 - Cos x.

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{2Sin^2\, \frac{x}{2}}{x^2}:⟹ </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>2Sin </p><p>2</p><p> </p><p>2</p><p>x$

$:\implies \displaystyle \tt\lim_{x \to 0} \frac{2Sin^2\, \frac{x}{2}}{4 (\frac{x}{2}) ^2}:⟹ </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>4( </p><p>2</p><p>x</p><p> </p><p> ) </p><p>2</p><p> </p><p>2Sin </p><p>2</p><p> </p><p>2</p><p>x</p><p> </p><p> </p><p> </p><p>$

Now, We know that,

$</p><p></p><p>\dagger \: \: \: \: \: \: \displaystyle \tt\lim_{x \to 0} \frac{Sin\,x}{x} = 1.† </p><p>x→0</p><p>lim</p><p>$