Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$
✓
Answer
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\text{ax}}{\text{x}}+\text{b}}{\text{a}+\frac{\sin\text{bx}}{\text{bx}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}\times\text{a}+\text{b}}{\text{a}+\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{bx}}{\text{bx}}\times\text{b}}$
$=\frac{\text{a}+\text{b}}{\text{a}+\text{b}}$
$=1$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\text{ax}}{\text{x}}+\text{b}}{\text{a}+\frac{\sin\text{bx}}{\text{bx}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}\times\text{a}+\text{b}}{\text{a}+\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{bx}}{\text{bx}}\times\text{b}}$
$=\frac{\text{a}+\text{b}}{\text{a}+\text{b}}$
$=1$
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