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Question Number 161538 by mnjuly1970 last updated on 19/Dec/21

$$$$$${lim}_{x\to -2}\left(\frac{2+3x+3x^2+x^3}{sin\left(\frac{\pi x}{2}\right)}\right)=?$$$$----$$

Answered by Ar Brandon last updated on 19/Dec/21

$$\mathcal{L}=\underset{x\to -2}{\lim }\left(\frac{1+{(1+x)}^3}{\sin \left(\frac{\pi x}{2}\right)}\right)=\underset{x\to -2}{\lim }\left(\frac{3{(1+x)}^2}{\frac{\pi }{2}\cos \left(\frac{\pi }{2}x\right)}\right)=-\frac{6}{\pi }$$

Commented by mnjuly1970 last updated on 19/Dec/21

$$thxsirbrandon..$$

Answered by cortano last updated on 19/Dec/21

$$\underset{x\to -2}{\lim }\left(\frac{x^3+3x^2+3x+2}{\sin \left(\frac{\pi x}{2}\right)}\right)$$$$=\underset{x\to -2}{\lim }\left(\frac{1+{(x+1)}^3}{\sin \left(\frac{\pi x}{2}\right)}\right)$$$$[letx+2=c]$$$$=\underset{c\to 0}{\lim }\frac{1+{(c-1)}^3}{\sin \frac{\pi }{2}(c-2)}$$$$=\underset{c\to 0}{\lim }\frac{1-{(1-c)}^3}{-\sin (\pi -\frac{\pi c}{2})}$$$$=\underset{c\to 0}{\lim }\frac{1-(1-3c)}{-\sin \frac{\pi c}{2}}=-\frac{6}{\pi }$$

Commented by mnjuly1970 last updated on 19/Dec/21

$$thankyousir..$$

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