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Exercise 12.1 · Q22

Q.lim⁡x→π2tan⁡2xx−π2\lim_{x\to \frac{\pi}{2}}\dfrac{\tan 2x}{x - \frac{\pi}{2}}

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This limit is a 0/0 indeterminate form that we resolve by rewriting tan⁡2x\tan 2x as sin⁡2xcos⁡2x\frac{\sin 2x}{\cos 2x} and then using the standard limit lim⁡t→0sin⁡tt=1\lim_{t \to 0} \frac{\sin t}{t} = 1 after a substitution. The value is 2\boxed{2}.

The core idea here is that when xx approaches π2\frac{\pi}{2}, both the numerator tan⁡2x\tan 2x and the denominator x−π2x - \frac{\pi}{2} go to zero. This is a classic 00\frac{0}{0} form, so we can't just plug in. The trick is to transform the expression into something involving sin⁡θθ\frac{\sin \theta}{\theta}, which has a known limit of 1 as θ→0\theta \to 0.

Why does this work? Because tan⁡2x\tan 2x is messy near π2\frac{\pi}{2}, but its sine part behaves nicely — the trouble comes from the cosine in the denominator of tan⁡\tan. By rewriting and then shifting the variable, we isolate the sine limit.

Let's work through it step by step.

  1. Check the form.

    As x→π2x \to \frac{\pi}{2}, we have 2x→π2x \to \pi, so tan⁡2x=sin⁡2xcos⁡2x\tan 2x = \frac{\sin 2x}{\cos 2x}.

    sin⁡2x→sin⁡π=0\sin 2x \to \sin \pi = 0, and cos⁡2x→cos⁡π=−1\cos 2x \to \cos \pi = -1, so tan⁡2x→0−1=0\tan 2x \to \frac{0}{-1} = 0.

    The denominator x−π2→0x - \frac{\pi}{2} \to 0. So indeed we have 00\frac{0}{0}.

  2. Rewrite the numerator.

    Write tan⁡2x=sin⁡2xcos⁡2x\tan 2x = \frac{\sin 2x}{\cos 2x}. The limit becomes:

lim⁡x→π2sin⁡2x(x−π2)cos⁡2x.\lim_{x\to \frac{\pi}{2}} \frac{\sin 2x}{(x - \frac{\pi}{2}) \cos 2x}.

  1. Make a substitution to center at zero.

    Let t=x−π2t = x - \frac{\pi}{2}. Then x=t+π2x = t + \frac{\pi}{2}, and as x→π2x \to \frac{\pi}{2}, t→0t \to 0.

    Now rewrite everything in terms of tt:

    • 2x=2(t+π2)=2t+π2x = 2\left(t + \frac{\pi}{2}\right) = 2t + \pi.
    • sin⁡2x=sin⁡(2t+π)=−sin⁡2t\sin 2x = \sin(2t + \pi) = -\sin 2t (since sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin \theta).
    • cos⁡2x=cos⁡(2t+π)=−cos⁡2t\cos 2x = \cos(2t + \pi) = -\cos 2t (since cos⁡(θ+π)=−cos⁡θ\cos(\theta + \pi) = -\cos \theta).
    • The denominator x−π2=tx - \frac{\pi}{2} = t.

    So the limit becomes:

lim⁡t→0−sin⁡2tt⋅(−cos⁡2t)=lim⁡t→0sin⁡2ttcos⁡2t.\lim_{t \to 0} \frac{-\sin 2t}{t \cdot (-\cos 2t)} = \lim_{t \to 0} \frac{\sin 2t}{t \cos 2t}.

The negatives cancel neatly.

  1. Separate into known limits. Write: sin⁡2ttcos⁡2t=sin⁡2tt⋅1cos⁡2t.…\frac{\sin 2t}{t \cos 2t} = \frac{\sin 2t}{t} \cdot \frac{1}{\cos 2t}. …

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