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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$
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Answer

$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$ If t = x - a Then, as x → a, t → 0 $\therefore\ \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\big(\text{a}\sin(\text{t}+\text{a})-(\text{t}+\text{a})\sin\text{a}\big)}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}\cot\text{t}-\text{t}\sin\text{a}-\text{a}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{a}}$ $=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}(\cos\text{t}-1)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\cos\text{a}+\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{t}\cos\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}+\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)}{\text{a}(\text{t}+\text{a})\text{t}}-\lim\limits_{\text{t}\rightarrow0}\frac{\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$ $=\frac{\text{a}\cos\text{a}}{\text{a}^2}+0-\frac{\sin\text{a}}{\text{a}^2}$ $=\frac{\text{a}\cos\text{a}-\sin\text{a}}{\text{a}^2}$

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