Limits and Derivatives: Tricky Questions Solved (7)

easy • 3 min read
Tags Limits And Derivatives

Question

Evaluate lim⁡x→0sin⁡3xtan⁡5x\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{\tan 5x}.

Solution — Step by Step

At x=0x = 0: sin⁡3(0)=0\sin 3(0) = 0, tan⁡5(0)=0\tan 5(0) = 0. So we have an indeterminate form 0/00/0.

Recall lim⁡x→0sin⁡xx=1\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1 and lim⁡x→0tan⁡xx=1\displaystyle\lim_{x \to 0} \frac{\tan x}{x} = 1.

Rewrite:

sin⁡3xtan⁡5x=sin⁡3x3x⋅5xtan⁡5x⋅3x5x=sin⁡3x3x⋅5xtan⁡5x⋅35\frac{\sin 3x}{\tan 5x} = \frac{\sin 3x}{3x} \cdot \frac{5x}{\tan 5x} \cdot \frac{3x}{5x} = \frac{\sin 3x}{3x} \cdot \frac{5x}{\tan 5x} \cdot \frac{3}{5}

As x→0x \to 0, sin⁡3x3x→1\dfrac{\sin 3x}{3x} \to 1 (let u=3xu = 3x, u→0u \to 0) and 5xtan⁡5x→1\dfrac{5x}{\tan 5x} \to 1 (let v=5xv = 5x, v→0v \to 0).

lim⁡x→0sin⁡3xtan⁡5x=1⋅1⋅35=35\lim_{x \to 0} \frac{\sin 3x}{\tan 5x} = 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}

35\boxed{\frac{3}{5}}

Why This Works

The trick is to introduce the variable matching sin⁡/x\sin/x form, even if it requires multiplying and dividing by the same constant. Once you have sin⁡(something)/(same something)\sin(\text{something})/(\text{same something}), the limit is 11.

This pattern handles a huge family of trig limits without needing L’Hôpital. The key recipe:

sin⁡(ax)sin⁡(bx)→ab,tan⁡(ax)tan⁡(bx)→ab,sin⁡(ax)tan⁡(bx)→ab\frac{\sin(ax)}{\sin(bx)} \to \frac{a}{b}, \quad \frac{\tan(ax)}{\tan(bx)} \to \frac{a}{b}, \quad \frac{\sin(ax)}{\tan(bx)} \to \frac{a}{b}

as x→0x \to 0.

lim⁡x→0sin⁡xx=1\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1

lim⁡x→0tan⁡xx=1\displaystyle\lim_{x \to 0} \frac{\tan x}{x} = 1

lim⁡x→01−cos⁡xx2=12\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

lim⁡x→0ex−1x=1\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = 1

lim⁡x→0ln⁡(1+x)x=1\displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1

lim⁡x→0(1+x)n−1x=n\displaystyle\lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n

Alternative Method

L’Hôpital’s rule. Differentiate top and bottom:

lim⁡x→03cos⁡3x5sec⁡25x=3⋅15⋅1=35✓\lim_{x \to 0} \frac{3\cos 3x}{5\sec^2 5x} = \frac{3 \cdot 1}{5 \cdot 1} = \frac{3}{5} \checkmark

Faster if you’ve memorised derivatives, but the standard-limits approach is what NCERT and CBSE expect.

For Class 11 CBSE, don’t use L’Hôpital — it’s not in the syllabus. Use standard limits and algebraic manipulation. JEE allows L’Hôpital, so by Class 12 you have both tools.

Common Mistake

Students write lim⁡x→0sin⁡3xtan⁡5x=sin⁡0tan⁡0=00\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{\tan 5x} = \dfrac{\sin 0}{\tan 0} = \dfrac{0}{0} and stop. That’s not the answer — 0/00/0 is indeterminate, meaning we need more work to find the actual limit.

Another error: assuming sin⁡(3x)≈3x\sin(3x) \approx 3x and tan⁡(5x)≈5x\tan(5x) \approx 5x “to lowest order” and writing the limit as 3x/5x=3/53x/5x = 3/5 without justification. The answer is right, but the working is sloppy. Always show the multiplication-and-division step explicitly for full marks.

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