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Compute the limit:
[tex]\large\begin{array}{l} \mathsf{L=\underset{x\to 0}{\ell im}~\dfrac{x^2\cdot cos\,x}{1-cos\,x}} \end{array}[/tex]
I'm not going to use the L'Hôpital's rule to compute this limit here. Instead, subtract x² from the numerator, and then add it back, so the expression doesn't change:
[tex]\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~\dfrac{-\,x^2+x^2\cdot cos\,x+x^2}{1-cos\,x}} \end{array}[/tex]
Split it into two fractions:
[tex]\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~\left[\dfrac{-\,x^2+x^2\cdot cos\,x}{1-cos\,x}+\dfrac{x^2}{1-cos\,x}\right]} \end{array}[/tex]
Take out the common factor – x², and you have
[tex]\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~\left[\dfrac{-\,x^2\cdot (1-cos\,x)}{1-cos\,x}+\dfrac{x^2}{1-cos\,x}\right]}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\left[-\,x^2+\dfrac{x^2}{1-cos\,x}\right]} \end{array}[/tex]
Now, multiply both the numerator and the denominator by (1 + cos x):
[tex]\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~\left[-\,x^2+\dfrac{x^2\cdot (1+cos\,x)}{(1-cos\,x)\cdot (1+cos\,x)}\right]}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\left[-\,x^2+\dfrac{x^2\cdot (1+cos\,x)}{1-cos^2\,x}\right]\qquad\quad (but~1-cos^2\,x=sin^2\,x)}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\left[-\,x^2+\dfrac{x^2\cdot (1+cos\,x)}{sin^2\,x}\right]}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\left[-\,x^2+\dfrac{x^2}{sin^2\,x}\cdot (1+cos\,x)\right]} \end{array}[/tex]
[tex]\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~\left[-\,x^2+\dfrac{1}{~\frac{sin^2\,x}{x^2}~}\cdot (1+cos\,x)\right]}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~\left[-\,x^2+\dfrac{1}{\left(\frac{sin\,x}{x}\right)^{\!2}}\cdot (1+cos\,x)\right]}\\\\\\ \vdots\qquad\textsf{(applying limit properties)}\\\\\\ \mathsf{=\underset{x\to 0}{\ell im}~(-\,x^2)+\underset{x\to 0}{\ell im}~\dfrac{1}{\left(\frac{sin\,x}{x}\right)^{\!2}}\cdot (1+cos\,x)}\\\\ \mathsf{=\underset{x\to 0}{\ell im}~(-\,x^2)+\dfrac{1}{\underset{x\to 0}{\ell im}\left(\frac{sin\,x}{x}\right)^{\!2}}\cdot \underset{x\to 0}{\ell im}~(1+cos\,x)} \end{array}[/tex]
[tex]\large\begin{array}{l} \mathsf{=\underset{x\to 0}{\ell im}~(-\,x^2)+\dfrac{1}{\left(\underset{x\to 0}{\ell im}~\frac{sin\,x}{x}\right)^{\!2}}\cdot \underset{x\to 0}{\ell im}~(1+cos\,x)\qquad\left(but~\underset{x\to 0}{\ell im}~\dfrac{sin\,x}{x}=1\right)}\\\\ \mathsf{=-\,0^2+\dfrac{1}{1^2}\cdot (1+cos\,0)}\\\\ \mathsf{=-\,0^2+\dfrac{1}{1^2}\cdot (1+1)}\\\\ \mathsf{=-\,0+\dfrac{1}{1}\cdot 2} \end{array}[/tex]
[tex]\large\begin{array}{l} \mathsf{=-\,0+ 2}\\\\ \mathsf{=2}\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{\underset{x\to 0}{\ell im}~\dfrac{x^2\cdot cos\,x}{1-cos\,x}=2} \end{array}}\quad\longleftarrow\quad\textsf{and there it is.} \end{array}[/tex]
I hope this helps. =)
Tags: limit l'hôpital rule fundamental trigonometric trig sine sin differential integral calculus
