Question

limit X tends to zero x square cos x upon 1 minus cos x?​

Answer 1

Step-by-step explanation:

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Answer 2

Given,

$\[f\left( x \right) = \frac{{{x^2}\cos x}}{{1 - \cos x}}\]$

Solution,

Formula used,$\[{\cos ^2}x + {\sin ^2}x = 1\]$

Formula used,$\[\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\]$

Apply the formula to the given equation.

$\[\begin{array}{l} \Rightarrow f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\cos x}}{{1 - \cos x}}\\ \Rightarrow f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\cos x}}{{1 - \cos x}} \times \frac{{1 + \cos x}}{{1 + \cos x}}\\ \Rightarrow f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\left( {\cos x} \right)\left( {1 + \cos x} \right)}}{{1 - {{\cos }^2}x}}\end{array}\]$

Solve further,

$\[\begin{array}{l} \Rightarrow f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\left( {\cos x} \right)\left( {1 + \cos x} \right)}}{{{{\sin }^2}x}}\\ \Rightarrow f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\left( {\cos x} \right)\left( {1 + \cos x} \right)}}{{\frac{{{{\sin }^2}x}}{{{x^2}}} \times {x^2}}}\end{array}\]$

Apply limits.

$\[\begin{array}{l} \Rightarrow f\left( x \right) = \frac{{\cos 0^\circ \left( {1 + \cos 0^\circ } \right)}}{1}\\ \Rightarrow f\left( x \right) = \frac{{1\left( {1 + 1} \right)}}{1}\\ \Rightarrow f\left( x \right) = 2\end{array}\]$

Hence, the value is $\[2.\]$