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NCERT Exemplar · Q47

Q.Evaluate lim⁡y→0(x+y)sec⁡(x+y)−xsec⁡xy\lim_{y \to 0} \dfrac{(x + y)\sec(x + y) - x\sec x}{y}.

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This limit is the derivative of f(x)=xsec⁡xf(x) = x\sec x with respect to xx, evaluated at the point xx. The answer is sec⁡x+xsec⁡xtan⁡x\sec x + x\sec x \tan x.

The expression you’ve written is the definition of the derivative of a function — but not of xsec⁡xx\sec x with respect to yy. Look carefully: the variable approaching zero is yy, and the numerator is f(x+y)−f(x)f(x+y) - f(x) where f(t)=tsec⁡tf(t) = t\sec t. So this is exactly

lim⁡y→0f(x+y)−f(x)y=f′(x).\lim_{y \to 0} \frac{f(x+y) - f(x)}{y} = f'(x).

That’s the core insight. Once you see that, the problem reduces to differentiating xsec⁡xx\sec x with respect to xx.


Why this works

The limit-of-polynomial idea extends to any differentiable function: if you have a function g(t)g(t) that is smooth at a point, then near that point it behaves like a linear function plus higher-order terms. The limit above isolates the coefficient of the linear term — that’s the derivative. Here g(t)=tsec⁡tg(t) = t\sec t is differentiable for all tt where sec⁡t\sec t is defined (i.e., t≠π2+nπt \neq \frac{\pi}{2} + n\pi), so the limit exists and equals g′(x)g'(x).


Step-by-step

  1. Recognise the derivative form

    The limit is lim⁡y→0f(x+y)−f(x)y\displaystyle \lim_{y \to 0} \frac{f(x+y) - f(x)}{y} with f(t)=tsec⁡tf(t) = t\sec t. By definition, this is f′(x)f'(x).

  2. Differentiate f(x)=xsec⁡xf(x) = x\sec x

    Use the product rule:

f′(x)=ddx(x)⋅sec⁡x+x⋅ddx(sec⁡x).f'(x) = \frac{d}{dx}(x) \cdot \sec x + x \cdot \frac{d}{dx}(\sec x).

We know ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x \tan x.

  1. Compute each piece

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