Evaluate $\lim_{x \to 0} \dfrac{\sin x - 2\sin 3x + \sin 5x}{x}$.

Evaluate limx → 0 (sin x - 2sin 3x + sin 5x)/(x). – Class 11 Mathematics

NCERT Exemplar · Q27

\lim_{x \to 0} \dfrac{\sin x - 2\sin 3x + \sin 5x}{x}

Assam AhsecShort· 3mImportance★★★★★est

70% · 122/175 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

This limit problem is solved by recognizing the indeterminate $\frac{0}{0}$ form and then applying the fundamental limit $\lim_{x \to 0} \frac{\sin kx}{x} = k$ to each term in the expression. The final value of the limit is $\boxed{0}$ .

When evaluating limits, especially those involving trigonometric functions as $x \to 0$ , our first step is always to try direct substitution. If this yields a finite, defined value, that's our answer. However, if it results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , it signals that further manipulation is required.

In this problem, the expression is $\dfrac{\sin x - 2\sin 3x + \sin 5x}{x}$ .

If we substitute $x=0$ directly:

Numerator: $\sin(0) - 2\sin(3 \cdot 0) + \sin(5 \cdot 0) = 0 - 2(0) + 0 = 0$ .

Denominator: $0$ .

So, we have the indeterminate form $\frac{0}{0}$ . This means we need to simplify or transform the expression before evaluating the limit.

The key concept here is the fundamental trigonometric limit:

\lim_{x \to 0} \frac{\sin x}{x} = 1

This limit is incredibly powerful and forms the basis for many other trigonometric limits. A direct extension of this is:

k

›Proof

To see why $\lim_{x \to 0} \frac{\sin kx}{x} = k$ :

Let $y = kx$ . As $x \to 0$ , $y \to 0$ .

Then, $\lim_{x \to 0} \frac{\sin kx}{x} = \lim_{y \to 0} \frac{\sin y}{y/k} = \lim_{y \to 0} k \cdot \frac{\sin y}{y}$ .

Since $k$ is a constant, we can pull it out of the limit:

$= k \lim_{y \to 0} \frac{\sin y}{y}$ .

We know $\lim_{y \to 0} \frac{\sin y}{y} = 1$ .

Therefore, $\lim_{x \to 0} \frac{\sin kx}{x} = k \cdot 1 = k$ .

Our strategy will be to break down the given complex fraction into a sum of simpler fractions, each matching the form $\frac{\sin kx}{x}$ , and then apply this standard limit.

Here's the step-by-step evaluation:

Rewrite the expression by distributing the denominator: The given limit is $\lim_{x \to 0} \dfrac{\sin x - 2\sin 3x + \sin 5x}{x}$ . We can split this into individual terms:

\lim_{x \to 0} \left( \frac{\sin x}{x} - \frac{2\sin 3x}{x} + \frac{\sin 5x}{x} \right)

Apply the limit properties: The limit of a sum/difference is the sum/difference of the limits (provided each individual limit exists). $\lim_{x \to 0} \frac{\sin x}{x} - \lim_{x \to 0} \frac{2\sin 3x}{x} + \lim_{x \to 0} \frac{\sin 5x}{x}$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ26 — Evaluate limx → 0 dfrac√(2) - √(1 + cos x)sin² x.→Next questionQ28 — If limx → 1 (x⁴ - 1)/(x - 1) = limx → k (x³ - k³)/(x² - k²), then find…

Q.Evaluate lim⁡x→0sin⁡x−2sin⁡3x+sin⁡5xx\lim_{x \to 0} \dfrac{\sin x - 2\sin 3x + \sin 5x}{x}limx→0​xsinx−2sin3x+sin5x​.

🎓Unlock everything free for 14 days

Q.Evaluate $\lim_{x \to 0} \dfrac{\sin x - 2\sin 3x + \sin 5x}{x}$ .

Unlock everything free for 14 days