Tough Limit? Use Standard Limits & Win! | JEE Main Maths ⚡

 

❓ Question

Evaluate the limit:

$$\underset{x\to 0^{+}}{\lim }\frac{\tan \text{\hspace{0.17em}⁣}\left(5x^{1\mathrm{/}3}\right)\ln \text{\hspace{0.17em}⁣}(1+3x^2)}{{({\tan }^{-1}(3\sqrt{x}))}^2[e^{5x^{4\mathrm{/}3}}-1]}$$

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🖼️ Question Image

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✍️ Short Solution

This is a standard JEE-type limit based on small-angle and small-x approximations.
As $x \to 0^+$, each function behaves like its first-order term.

We’ll convert every function into its leading approximation and simplify.

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🔹 Step 1 — Recall standard limits

As $t \to 0$:

$$\tan t \sim t$$
$$\ln(1+t) \sim t$$
$$\tan^{-1} t \sim t$$
$$e^t - 1 \sim t$$

These shortcuts are mandatory for fast JEE solving.

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🔹 Step 2 — Apply approximations one by one

1️⃣ Tangent term

$$\tan(5x^{1/3}) \sim 5x^{1/3}$$

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2️⃣ Logarithmic term

$$\ln(1 + 3x^2) \sim 3x^2$$

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3️⃣ Inverse tangent term

$${\tan }^{-1}(3\sqrt{x})\sim 3\sqrt{x}$$

So,

$$({\tan }^{-1}(3\sqrt{x}){)}^2\sim (3\sqrt{x}{)}^2=9x$$

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4️⃣ Exponential term

$$e^{5x^{4/3}} - 1 \sim 5x^{4/3}$$

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🔹 Step 3 — Substitute approximations into the limit

Numerator becomes:

$$(5x^{1/3})(3x^2) = 15x^{7/3}$$

Denominator becomes:

$$(9x)(5x^{4/3}) = 45x^{7/3}$$

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🔹 Step 4 — Simplify

$$\frac{15x^{7/3}}{45x^{7/3}} = \frac{15}{45} = \frac{1}{3}$$

All powers of $x$ cancel perfectly — a strong check that the method is correct.

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✅ Final Answer

$$\boxed{{\displaystyle \frac{1}{\mathrm{3<span\; class="vlist-s"></span>}}}}$$

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