Limits and Derivatives: Step-by-Step Worked Examples (8)

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Tags Limits And Derivatives

Question

Evaluate the limit:

lim⁡x→0sin⁡5x−sin⁡3xx\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{x}

Then differentiate f(x)=x2sin⁡xf(x) = x^2 \sin x from first principles.

Solution — Step by Step

lim⁡x→0sin⁡5x−sin⁡3xx=lim⁡x→0[sin⁡5xx−sin⁡3xx]\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{x} = \lim_{x \to 0} \left[\frac{\sin 5x}{x} - \frac{\sin 3x}{x}\right]

We know lim⁡x→0sin⁡kxx=k\lim_{x \to 0} \frac{\sin kx}{x} = k. So:

=5−3=2= 5 - 3 = 2

First answer: limit = 2.

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0(x+h)2sin⁡(x+h)−x2sin⁡xhf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{(x + h)^2 \sin(x + h) - x^2 \sin x}{h}

Let A=(x+h)2sin⁡(x+h)A = (x + h)^2 \sin(x + h) and B=x2sin⁡xB = x^2 \sin x. Insert ±(x+h)2sin⁡x\pm (x + h)^2 \sin x:

A−B=(x+h)2[sin⁡(x+h)−sin⁡x]+[(x+h)2−x2]sin⁡xA - B = (x + h)^2[\sin(x + h) - \sin x] + [(x + h)^2 - x^2]\sin x

f′(x)=lim⁡h→0(x+h)2⋅sin⁡(x+h)−sin⁡xh+lim⁡h→0(x+h)2−x2hsin⁡xf'(x) = \lim_{h \to 0} (x + h)^2 \cdot \frac{\sin(x + h) - \sin x}{h} + \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} \sin x

The first limit: (x+h)2→x2(x + h)^2 \to x^2 and sin⁡(x+h)−sin⁡xh→cos⁡x\frac{\sin(x + h) - \sin x}{h} \to \cos x (definition of derivative of sin⁡\sin).

The second limit: (x+h)2−x2h=2x+h→2x\frac{(x + h)^2 - x^2}{h} = 2x + h \to 2x.

f′(x)=x2cos⁡x+2xsin⁡xf'(x) = x^2 \cos x + 2x \sin x

Second answer: f′(x)=x2cos⁡x+2xsin⁡xf'(x) = x^2 \cos x + 2x \sin x.

Why This Works

The “split the fraction” trick converts an indeterminate 0/00/0 form into two known standard limits. The sin⁡kx/x→k\sin kx / x \to k rule comes directly from sin⁡kx≈kx\sin kx \approx kx for small xx.

For first-principles differentiation, the add-and-subtract trick produces the product rule organically. We see f′(x)=u′v+uv′f'(x) = u' v + u v' emerging from the algebra without invoking the rule by name.

Alternative Method

For the limit, use the sum-to-product identity:

sin⁡5x−sin⁡3x=2cos⁡4xsin⁡x\sin 5x - \sin 3x = 2 \cos 4x \sin x

So the limit becomes 2cos⁡4x⋅sin⁡xx→2⋅1⋅1=22 \cos 4x \cdot \frac{\sin x}{x} \to 2 \cdot 1 \cdot 1 = 2.

For the derivative, apply the product rule directly: (x2)′sin⁡x+x2(sin⁡x)′=2xsin⁡x+x2cos⁡x(x^2)' \sin x + x^2 (\sin x)' = 2x \sin x + x^2 \cos x.

Common Mistake

Students apply lim⁡sin⁡x/x=1\lim \sin x / x = 1 when the argument is not the same as the denominator. For sin⁡5x/x\sin 5x / x, the rule gives 55, not 11. The numerator and denominator must be matched (or you must adjust by multiplying and dividing by the right factor).

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