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Exercise 12.1 · Q20

Q.lim⁡x→0sin⁡ax+bxax+sin⁡bx, a,b,a+b≠0\lim_{x\to 0}\dfrac{\sin ax + bx}{ax + \sin bx},\ a, b, a + b \neq 0

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The limit is 11 because both numerator and denominator behave like (a+b)x(a+b)x near x=0x=0, using the standard limit lim⁡x→0sin⁡xx=1\lim_{x\to 0} \frac{\sin x}{x} = 1.

Why this works

When xx is very close to 00, the sine function behaves almost like its argument: sin⁡θ≈θ\sin \theta \approx \theta. This is the core idea behind the limit lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1. For a polynomial-like combination of sines and linear terms, we can rewrite each sine as its argument times a correction factor that tends to 11, then simplify.

The given expression is a ratio of two sums, each containing a sine term and a linear term. The trick is to factor out xx from both numerator and denominator, then use the standard sine limit on the sin⁡(ax)ax\frac{\sin(ax)}{ax} and sin⁡(bx)bx\frac{\sin(bx)}{bx} pieces.

Watch out

A common mistake is to replace sin⁡(ax)\sin(ax) with axax directly and cancel everything, forgetting that the limit is taken as x→0x \to 0, not at x=0x=0. The correct approach is to factor and use the limit form, not to substitute x=0x=0 into the original expression (which gives 00\frac{0}{0}).

Step-by-step solution

  1. Rewrite each sine term using the standard limit. We know lim⁡x→0sin⁡(ax)ax=1\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1 and lim⁡x→0sin⁡(bx)bx=1\lim_{x \to 0} \frac{\sin(bx)}{bx} = 1. So we can write:

sin⁡(ax)=ax⋅sin⁡(ax)ax,sin⁡(bx)=bx⋅sin⁡(bx)bx\sin(ax) = ax \cdot \frac{\sin(ax)}{ax}, \quad \sin(bx) = bx \cdot \frac{\sin(bx)}{bx}

where each fraction tends to 11 as x→0x \to 0.

  1. Factor xx from numerator and denominator.

    Numerator: sin⁡(ax)+bx=ax⋅sin⁡(ax)ax+bx=x(a⋅sin⁡(ax)ax+b)\sin(ax) + bx = ax \cdot \frac{\sin(ax)}{ax} + bx = x\left( a \cdot \frac{\sin(ax)}{ax} + b \right)

    Denominator: ax+sin⁡(bx)=ax+bx⋅sin⁡(bx)bx=x(a+b⋅sin⁡(bx)bx)ax + \sin(bx) = ax + bx \cdot \frac{\sin(bx)}{bx} = x\left( a + b \cdot \frac{\sin(bx)}{bx} \right)

  2. Cancel the common factor xx. …

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