Question

Evaluate the limit of (tanx)^2/(1-cosx) as x tends to 0
Please show your work. Thanks!

Answer 1

Rewrite the limand as

$\dfrac{\tan^2(x)}{1 - \cos(x)} = \dfrac{\sin^2(x)}{\cos^2(x)(1 - \cos(x))}$

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

$\dfrac{\sin^2(x)}{\cos^2(x)(1 - \cos(x))} \cdot \dfrac{1+\cos(x)}{1+\cos(x)} = \dfrac{\sin^2(x)(1 + \cos(x))}{\cos^2(x)(1 - \cos^2(x))} \\\\= \dfrac{\sin^2(x)(1 + \cos(x))}{\cos^2(x)\sin^2(x)} \\\\ = \dfrac{1 + \cos(x)}{\cos^2(x)}$

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

Answer 2

We are given with a limit :-

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

We know that ;

• ${\boxed{\bf \dfrac{\sin (\theta)}{\cos (\theta)}=\tan (\theta)}}$

Using this , the above limit can be written as ;

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

Multiplying and dividing both numerator and denominator by ${\sf 1+\cos (x)}$

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

We knows a identity that is

• ${\boxed{\bf a^{2}-b^{2}=(a+b)(a-b)}}$

Using this we have ;

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

We knows another Trigonometric identity i.e ;

• ${\boxed{\bf 1-\cos^{2}(\theta)=\sin^{2}(\theta)}}$

Now , using this we have ;

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

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

Putting the limit we have ;

${:\implies \quad \displaystyle \sf \dfrac{1+\cos (0)}{\cos^{2}(0)}}$

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

${:\implies \quad \displaystyle \sf \dfrac{2}{1}}$

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