Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$
✓
Answer
$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$ $\Rightarrow\text{x}\rightarrow1,\text{x}-1\rightarrow0,$ let $\text{x}-1=\text{y}\Rightarrow\text{y}\rightarrow0$ $=\lim\limits_{\text{x}\rightarrow{0}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}=\lim\limits_{\text{x}-1\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin\pi\text{ x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{(-\text{y})(1+\text{y}+1)}{\sin\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin\pi(\text{y}+1)}$ $=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{-\sin\pi(\text{y}+1)}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\frac{\sin\pi\text{y}}{2}}$ $=\frac{\lim\limits_{\text{y}\rightarrow{0}}\text{y}(\text{y}+2)}{\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\pi\text{y}}{\pi\text{y}}\times\pi\text{y}}$ $=\frac{2}{\pi}$
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