Evaluate $\lim_{x \to 0} \dfrac{\sin 2x + 3x}{2x + \tan 3x}$.

> [!TLDR] > This limit is a $\frac{0}{0}$ form. By factoring $x$ out of numerator and denominator, then using the standard limits $\lim_{x\to 0} \frac{\sin x}{x}=1$ and $\lim_{x\to 0} \frac{\tan x}{x}=1$, the limit simplifies to $\frac{5}{5}=1$. The key idea: when $x \to 0$, both $\sin 2x$ and $\tan 3x$ behave like $2x$ and $3x$ respectively. So the whole expression should approach $\frac{2x+3x}{2x+3x} = 1$. But we need to prove it cleanly. **Why this approach works** — The limit of a polynomial (or a ratio of polynomials in $\sin$ and $\tan$) as $x \to 0$ can often be handled by factoring out the smallest power of $x$ from each term. Here, every term contains $x$ as a factor (since $\sin 2x \approx 2x$, $\tan 3x \approx 3x$). So we factor $x$ out of numerator and denominator, cancel it, and then use the standard trigonometric limits. Let’s work through it step by step. 1. **Check the form.** As $x \to 0$, $\sin 2x \to 0$, $3x \to 0$, so numerator $\to 0$. Denominator: $2x \to 0$, $\tan 3x \to 0$, so denominator $\to 0$. This is a $\frac{0}{0}$ indeterminate form — we need to simplify. 2. **Factor $x$ from numerator and denominator.** Numerator: $\sin 2x + 3x = x\left(\frac{\sin 2x}{x} + 3\right)$ Denominator: $2x + \tan 3x = x\left(2 + \frac{\tan 3x}{x}\right)$ So the limit becomes: \[ \lim_{x \to 0} \frac{x\left(\frac{\sin 2x}{x} + 3\right)}{x\left(2 + \frac{\tan 3x}{x}\right)} = \lim_{x \to 0} \frac{\frac{\sin 2x}{x} + 3}{2 + \frac{\tan 3x}{x}} \] The $x$ cancels safely because $x \neq 0$ in the limit process. 3. **Rewrite the trigonometric ratios to use standard limits.** \[ \frac{\sin 2x}{x} = 2 \cdot \frac{\sin 2x}{2x} \] \[ \frac{\tan 3x}{x} = 3 \cdot \frac{\tan 3x}{3x} \] Now the limit is: \[ \lim_{x \to 0} \frac{2 \cdot \frac{\sin 2x}{2x} + 3}{2 + 3 \cdot \frac{\tan 3x}{3x}} \] 4. **Apply the standard limits.** We know: \[ \lim_{u \to 0} \frac{\sin u}{u} = 1 \quad \text{and} \quad

Evaluate limx → 0 (sin 2x + 3x)/(2x + tan 3x).

NCERT Exemplar · Q23

\lim_{x \to 0} \dfrac{\sin 2x + 3x}{2x + \tan 3x}

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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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This limit is a $\frac{0}{0}$ form. By factoring $x$ out of numerator and denominator, then using the standard limits $\lim_{x\to 0} \frac{\sin x}{x}=1$ and $\lim_{x\to 0} \frac{\tan x}{x}=1$ , the limit simplifies to $\frac{5}{5}=1$ .

The key idea: when $x \to 0$ , both $\sin 2x$ and $\tan 3x$ behave like $2x$ and $3x$ respectively. So the whole expression should approach $\frac{2x+3x}{2x+3x} = 1$ . But we need to prove it cleanly.

Why this approach works — The limit of a polynomial (or a ratio of polynomials in $\sin$ and $\tan$ ) as $x \to 0$ can often be handled by factoring out the smallest power of $x$ from each term. Here, every term contains $x$ as a factor (since $\sin 2x \approx 2x$ , $\tan 3x \approx 3x$ ). So we factor $x$ out of numerator and denominator, cancel it, and then use the standard trigonometric limits.

Let’s work through it step by step.

Check the form. As $x \to 0$ , $\sin 2x \to 0$ , $3x \to 0$ , so numerator $\to 0$ . Denominator: $2x \to 0$ , $\tan 3x \to 0$ , so denominator $\to 0$ . This is a $\frac{0}{0}$ indeterminate form — we need to simplify.
Factor $x$ from numerator and denominator.

Numerator: $\sin 2x + 3x = x\left(\frac{\sin 2x}{x} + 3\right)$

Denominator: $2x + \tan 3x = x\left(2 + \frac{\tan 3x}{x}\right)$

So the limit becomes:

\lim_{x \to 0} \frac{x\left(\frac{\sin 2x}{x} + 3\right)}{x\left(2 + \frac{\tan 3x}{x}\right)} = \lim_{x \to 0} \frac{\frac{\sin 2x}{x} + 3}{2 + \frac{\tan 3x}{x}}

The $x$ cancels safely because $x \neq 0$ in the limit process.

Rewrite the trigonometric ratios to use standard limits.

\frac{\sin 2x}{x} = 2 \cdot \frac{\sin 2x}{2x}

\frac{\tan 3x}{x} = 3 \cdot \frac{\tan 3x}{3x}

Now the limit is:

\lim_{x \to 0} \frac{2 \cdot \frac{\sin 2x}{2x} + 3}{2 + 3 \cdot \frac{\tan 3x}{3x}}

Apply the standard limits. We know:

\lim_{u \to 0} \frac{\sin u}{u} = 1 \quad \text{and} \quad \lim_{u \to 0} \frac{\tan u}{u} = 1

So as $x \to 0$ , $2x \to 0$ and $3x \to 0$ , giving:

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