If $f(x) = \dfrac{\tan x}{x - \pi}$, then $\lim_{x \to \pi} f(x) = $ ________.

> [!TLDR] > The limit is found by rewriting the expression so that the denominator’s zero is cancelled by a matching zero in the numerator. Using the substitution $t = x - \pi$ and the small-angle approximation $\tan t \approx t$, the limit evaluates to $1$. The core idea here is that when you plug $x = \pi$ directly into $f(x)$, you get $\frac{0}{0}$ — an indeterminate form. That tells us there might be a hidden cancellation. The trick is to shift the variable so the troublesome point moves to $0$, where limits of trigonometric functions are well-known. --- 1. **Substitute to shift the limit point** Let $t = x - \pi$. Then as $x \to \pi$, we have $t \to 0$. Also $x = t + \pi$, so \[ f(x) = \frac{\tan(t + \pi)}{t}. \] Now recall the periodicity of tangent: $\tan(\theta + \pi) = \tan \theta$. So \[ \tan(t + \pi) = \tan t. \] Therefore \[ f(x) = \frac{\tan t}{t}, \quad \text{with } t \to 0. \] 2. **Use the fundamental limit for $\tan$** We know that \[ \lim_{t \to 0} \frac{\tan t}{t} = 1. \] Why? Because $\tan t = \frac{\sin t}{\cos t}$, so \[ \frac{\tan t}{t} = \frac{\sin t}{t} \cdot \frac{1}{\cos t}. \] As $t \to 0$, $\frac{\sin t}{t} \to 1$ and $\cos t \to 1$, so the product tends to $1$. > [!FORMULA] > \[ > \lim_{t \to 0} \frac{\tan t}{t} = 1 > \] 3. **Apply the limit** Since $\frac{\tan t}{t} \to 1$ as $t \to 0$, we have \[ \lim_{x \to \pi} f(x) = \lim_{t \to 0} \frac{\tan t}{t} = 1. \] …

If f(x) = (tan x)/(x - π), then limx → π f(x) = __ – Class 11 Mathematics

NCERT Exemplar · Q77

f(x) = \dfrac{\tan x}{x - \pi}

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Concept understanding — Standard Limits

A Toolkit of Named Limits

Some limits recur so often across problems that it is worth memorising their values outright, along with the one theorem that proves the trickiest of them: the Sandwich (Squeeze) Theorem.

The Sandwich Theorem

Theorem 9.5. If $g(x) \le f(x) \le h(x)$ for all $x$ near $x_0$ (except possibly at $x_0$ itself), and if

\lim_{x\to x_0} g(x) = \lim_{x\to x_0} h(x) = l,

then $\lim_{x\to x_0} f(x) = l$ too — $f$ is "squeezed" between two functions that agree in the limit, so it has no room to do anything else.

Illustration: to show $\displaystyle\lim_{x\to 0} x^2\sin\!\frac{1}{x^2} = 0$ , note that $\sin(\cdot)$ is always between $-1$ and $1$ , so $-x^2 \le x^2\sin\frac{1}{x^2} \le x^2$ . Since both $-x^2$ and $x^2$ tend to $0$ as $x\to 0$ , the Sandwich Theorem forces the middle expression to $0$ as well — even though $\lim_{x\to 0}\sin\frac{1}{x^2}$ on its own does not exist (it oscillates wildly), so the product rule alone could never have been applied directly.

Watch out

This is exactly why the Sandwich Theorem is indispensable rather than a curiosity: whenever one factor oscillates without a limit but is bounded, and the other factor is squeezed to zero, the ordinary product law (Concept 2) is not applicable — you need the sandwich.

The two flagship trigonometric limits

Result 9.1.

\text{(a)}\quad \lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1 \qquad\qquad \text{(b)}\quad \lim_{\theta\to 0}\frac{1-\cos\theta}{\theta} = 0

Part (a) is proved geometrically by sandwiching $\frac{\sin\theta}{\theta}$ between $\cos\theta$ and $1$ using the areas of a triangle, a sector, and a larger triangle built on the unit circle; since both bounding functions tend to $1$ as $\theta\to0$ , so must $\frac{\sin\theta}{\theta}$ . Part (b) follows algebraically from (a) by writing $1-\cos\theta = 2\sin^2\frac{\theta}{2}$ and splitting the quotient into a $\sin$ -over-argument piece (which uses part (a)) times a factor that vanishes.

A direct corollary worth keeping separate: $\displaystyle\lim_{x\to 0}\sin x = 0$ , obtained from the sandwich $-|x|\le \sin x \le |x|$ .

The full standard-limit toolkit (§9.2.10)

Alongside the trig pair above, these are worth having on instant recall — none require anything beyond algebra and substitution to use (their proofs, where given, lean on the exponential/log relationship or on the trig pair):

\lim_{x\to 0}\frac{e^x-1}{x}=1 \qquad\quad \lim_{x\to 0}\frac{a^x-1}{x}=\log a \ (a>0) \qquad\quad \lim_{x\to 0}\frac{\log(1+x)}{x}=1

\lim_{x\to 0}\frac{\sin^{-1}x}{x}=1 \qquad\quad \lim_{x\to 0}\frac{\tan^{-1}x}{x}=1

And the three equivalent forms of the number $e$ as a limit:

\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e \qquad\quad \lim_{x\to 0}(1+x)^{1/x} = e \qquad\quad \lim_{x\to\infty}\left(1+\frac{k}{x}\right)^x = e^k

Note

$e$ is a transcendental number — it never satisfies any polynomial equation with rational coefficients. That's part of why it shows up as a genuinely new limiting constant here rather than something expressible in simpler closed form.

The recognise-and-substitute pattern

Nearly every "hard-looking" limit in this section is really one of the above standard forms in disguise, reached via a clean substitution $y = (\text{some expression in } x)$ chosen so that $y \to 0$ (or $y\to\infty$ ) exactly when $x$ does. The book's worked examples all follow this shape:

Spot the shell. Identify which standard form the expression resembles — a $\left(1+\Box\right)^{1/\Box}$ shape signals $e$ ; a $\dfrac{\sin(\Box)}{\Box}$ shape signals Result 9.1(a); a $\dfrac{a^{\Box}-1}{\Box}$ shape signals Result 9.3.
Substitute $y$ for the " $\Box$ " so the expression matches the standard form exactly, tracking what $y \to$ as $x \to x_0$ .
Apply the standard limit to the $y$ -expression, then (if the exponent or coefficient outside doesn't vanish) combine using the power/product rules from Concept 2. …

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The limit is found by rewriting the expression so that the denominator’s zero is cancelled by a matching zero in the numerator. Using the substitution $t = x - \pi$ and the small-angle approximation $\tan t \approx t$ , the limit evaluates to $1$ .

The core idea here is that when you plug $x = \pi$ directly into $f(x)$ , you get $\frac{0}{0}$ — an indeterminate form. That tells us there might be a hidden cancellation. The trick is to shift the variable so the troublesome point moves to $0$ , where limits of trigonometric functions are well-known.

Substitute to shift the limit point Let $t = x - \pi$ . Then as $x \to \pi$ , we have $t \to 0$ . Also $x = t + \pi$ , so

f(x) = \frac{\tan(t + \pi)}{t}.

Now recall the periodicity of tangent: $\tan(\theta + \pi) = \tan \theta$ . So

\tan(t + \pi) = \tan t.

Therefore

f(x) = \frac{\tan t}{t}, \quad \text{with } t \to 0.

Use the fundamental limit for $\tan$ We know that

\lim_{t \to 0} \frac{\tan t}{t} = 1.

Why? Because $\tan t = \frac{\sin t}{\cos t}$ , so

\frac{\tan t}{t} = \frac{\sin t}{t} \cdot \frac{1}{\cos t}.

As $t \to 0$ , $\frac{\sin t}{t} \to 1$ and $\cos t \to 1$ , so the product tends to $1$ .

\lim_{t \to 0} \frac{\tan t}{t} = 1

Apply the limit Since $\frac{\tan t}{t} \to 1$ as $t \to 0$ , we have $\lim_{x \to \pi} f(x) = \lim_{t \to 0} \frac{\tan t}{t} = 1.$ …

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Q.If f(x)=tan⁡xx−πf(x) = \dfrac{\tan x}{x - \pi}f(x)=x−πtanx​, then lim⁡x→πf(x)=\lim_{x \to \pi} f(x) = limx→π​f(x)= ________.

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Q.If $f(x) = \dfrac{\tan x}{x - \pi}$ , then $\lim_{x \to \pi} f(x) =$ ________.

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