Question

lim 1+cos^3 θ/sin ^2 θ
θ→π
​ ​

Answer 1

Question :-

Evaluate

$\rm \: \displaystyle\lim_{\theta \to \pi}\rm \frac{1 + {cos}^{3} \theta }{ {sin}^{2}\theta }$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{\theta \to \pi}\rm \frac{1 + {cos}^{3} \theta }{ {sin}^{2}\theta }$

If we substitute directly, we get

$\rm \: = \: \dfrac{1 + {cos}^{3} \pi }{ {sin}^{2}\pi }$

We know,

$\boxed{\sf{  \:\rm \: cos\pi = - 1 \: }} \: \: \rm \: \: and \: \: \: \: \boxed{\sf{  \:\rm \: sin\pi = 0 \: }}$

So, on substituting these values, we get

$\rm \: =  \: \dfrac{1 - 1}{0}$

$\rm \: =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{\theta \to \pi}\rm \frac{1 + {cos}^{3} \theta }{ {sin}^{2}\theta }$

We know,

$\boxed{\sf{  \:\rm \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this, we get

$\rm \: =  \: \displaystyle\lim_{\theta \to \pi}\rm \frac{1 + {cos}^{3} \theta }{1 - {cos}^{2}\theta }$

We know,

$\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}$

and

$\boxed{\sf{  \:\rm \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy) \: }}$

So, using these results, we get

$\rm \: =  \: \displaystyle\lim_{\theta \to \pi}\rm \frac{ \cancel{(1 + cos\theta )} \: \: ( {1}^{2} + {cos}^{2}\theta - cos\theta \times 1) }{ \cancel{(1 + cos\theta )} \: \: (1 - cos\theta )}$

$\rm \: =  \: \displaystyle\lim_{\theta \to \pi}\rm \frac{1 + {cos}^{2} \theta - cos\theta }{1 - cos\theta }$

$\rm \: =  \: \dfrac{1 + {( - 1)}^{2} - ( - 1)}{1 - ( - 1)}$

$\rm \: =  \: \frac{1 + 1 + 1}{1 + 1}$

$\rm \: =  \: \dfrac{3}{2}$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \displaystyle\lim_{\theta \to \pi}\rm \frac{1 + {cos}^{3} \theta }{ {sin}^{2}\theta } = \frac{3}{2} \: \: }}$

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Additional Information :-

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1 \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga \: }}$