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LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES4 Marks
Question
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$At x = 0, the value of the given function takes the form$\frac{0}{0}$.
$\text{Now},$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}=\frac{1}{\text{b}}\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{a}+\cos\text{x})}{\sin\text{x}}$
$=\frac{1}{\text{b}}\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{\text{x}}{\sin\text{x}}\bigg)\times\lim\limits_{\text{x}\rightarrow0}(\text{a}+\cos\text{x})$
$=\frac{1}{\text{b}}\times\bigg(\frac{1}{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}\bigg) \times\lim\limits_{\text{x}\rightarrow0}{(\text{a}+\cos\text{x})}$
$=\frac{1}{\text{b}}\times(\text{a}+\cos0) \bigg[\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\bigg]$
$=\frac{\text{a}+1}{\text{b}}$

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