CUET UG 2025 — Mathematics Calculus

For the function to be continuous at $x = 0$x=0, we need:

$$\lim_{x \to 0} f(x) = f(0)$$x→0lim​f(x)=f(0)

From the definition, $f(0) = k$f(0)=k.

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So we need to find:

$$\lim_{x \to 0} \dfrac{3x + 4\tan x}{x}$$x→0lim​x3x+4tanx​

Splitting the fraction by dividing each term in the numerator by $x$x separately:

$$= \lim_{x \to 0} \left[\dfrac{3x}{x} + \dfrac{4\tan x}{x}\right]$$=x→0lim​[x3x​+x4tanx​]

$$= \lim_{x \to 0} 3 + \lim_{x \to 0} 4 \cdot \dfrac{\tan x}{x}$$=x→0lim​3+x→0lim​4⋅xtanx​

We can split the limit like this because each individual limit exists and is finite.

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Using the standard result $\displaystyle \lim_{x \to 0} \dfrac{\tan x}{x} = 1$x→0lim​xtanx​=1 :

This comes from the fact that $\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}$xtanx​=xsinx​⋅cosx1​, and since $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$x→0lim​xsinx​=1 and $\displaystyle \lim_{x \to 0} \dfrac{1}{\cos x} = 1$x→0lim​cosx1​=1, the result follows.

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Substituting back:

$$= 3 + 4 \cdot 1$$=3+4⋅1

$$= 3 + 4 = 7$$=3+4=7

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Applying the continuity condition:

$$\lim_{x \to 0} f(x) = f(0)$$x→0lim​f(x)=f(0)

$$7 = k$$7=k

$$\boxed{k = 7}$$k=7​