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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x}+\text{a}\sin\text{bx})^\frac{1}{\text{x}}$
✓

Answer

$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x}+\text{a}\sin\text{bx})^\frac{1}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}(1+(\cos\text{x}+\text{a}\sin\text{bx}))^\frac{1}{\text{x}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{(\cos\text{x}+\text{a}\sin\text{bx}-1)}{\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a}\sin\text{bx}-(1-\cos\text{x}))}{\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{a}\sin\text{bx}-2\sin^2\big(\frac{\text{x}}{2}\big)\big)}{\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{\text{ab}\sin\text{bx}}{\text{bx}}-\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{x}}{2}\big)\sin\big(\frac{\text{x}}{2}\big)}{2\big(\frac{\text{x}}{2}\big)}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{\text{ab}\sin\text{bx}}{\text{bx}}-\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{x}}{2}\big)\sin\big(\frac{\text{x}}{2}\big)}{\big(\frac{\text{x}}{2}\big)}}$
$=\text{e}^{\text{ab}-0}$
$=\text{e}^{\text{ab}}$

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