Respuesta :

Rewrite the limand as

[tex]\dfrac{\tan^2(x)}{1 - \cos(x)} = \dfrac{\sin^2(x)}{\cos^2(x)(1 - \cos(x))}[/tex]

Multiply the numerator and denominator by 1 + cos(x). Then

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

As x goes to 0, the limand goes to (1 + 1)/1² = 2.

We are given with a limit :-

[tex]{\quad \blacktriangleright \blacktriangleright \blacktriangleright {\displaystyle \sf \lim_{x\to 0}\dfrac{\tan^{2}(x)}{1-\cos (x)}}}[/tex]

We know that ;

  • [tex]{\boxed{\bf \dfrac{\sin (\theta)}{\cos (\theta)}=\tan (\theta)}}[/tex]

Using this , the above limit can be written as ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}\dfrac{\sin^{2}(x)}{\cos^{2}(x)\{1-\cos (x)\}}}[/tex]

Multiplying and dividing both numerator and denominator by [tex]{\sf 1+\cos (x)}[/tex]

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}\dfrac{\sin^{2}(x)\{1+\cos (x)\}}{\cos^{2}(x)\{1-\cos (x)\}\{1+\cos (x)\}}}[/tex]

We knows a identity that is

  • [tex]{\boxed{\bf a^{2}-b^{2}=(a+b)(a-b)}}[/tex]

Using this we have ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}\dfrac{\sin^{2}(x)\{1+\cos (x)\}}{\cos^{2}(x)\{1-\cos^{2}(x)\}}}[/tex]

We knows another Trigonometric identity i.e ;

  • [tex]{\boxed{\bf 1-\cos^{2}(\theta)=\sin^{2}(\theta)}}[/tex]

Now , using this we have ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}\dfrac{\cancel{\sin^{2}(x)}\{1+\cos (x)\}}{\cos^{2}(x)\cancel{\sin^{2}(x)}}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}\dfrac{1+\cos (x)}{\cos^{2}(x)}}[/tex]

Putting the limit we have ;

[tex]{:\implies \quad \displaystyle \sf \dfrac{1+\cos (0)}{\cos^{2}(0)}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{1+1}{1^{2}} \quad \qquad \{\because \cos (0)=1\}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{2}{1}}[/tex]

[tex]{:\implies \quad {\displaystyle \therefore \quad \bf \underline{\underline{\lim_{x\to 0}\dfrac{\tan^{2}(x)}{1-\cos (x)}=2}}}}[/tex]

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