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Mathematics · JEE

Limits, continuity and differentiability Short Tricks for JEE

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Quick answer

Short tricks for Limits, continuity and differentiability work only with strong fundamentals. Apply the tips below in timed sets and review every explanation.

Use these Limits, continuity and differentiability shortcuts to save time in JEE Mathematics papers — then validate speed with 28+ MCQs on Goodmarks.

Short tricks for speed

  • Limits, continuity and differentiability focus drill

    Solve 15 mixed MCQs for Limits, continuity and differentiability, review every explanation, and note formulas you hesitated on.

  • Calculus substitution scan

    Spot standard forms (sin²x, 1/(a²+x²), e^ax) before integrating — JEE rewards pattern recognition.

  • Graph sketch shortcut

    For coordinate geometry, mark intercepts and asymptotes first; many MCQs need only qualitative graph features.

Syllabus context

Part of Limit, Continuity and Differentiability in JEE Main Mathematics.

Free sample questions

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Q1MathsUnit 7: Limit, Continuity and Differentiability
Let f(x)=x3−x2+x+1\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}^{3}-\boldsymbol{x}^{2}+\boldsymbol{x}+\mathbf{1} and g(x)=\boldsymbol{g}(\boldsymbol{x})= {max⁡{f(t)},0≤t≤x0≤x≤13−x,1<x≤2\left\{\begin{array}{l}\max \{f(t)\}, \quad 0 \leq t \leq x \quad 0 \leq x \leq 1 \\ 3-x, \quad 1<x \leq 2\end{array}\right. Then in the interval [0,2],g(x)[0,2], g(x) is This question has multiple correct options
Q2MathsUnit 7: Limit, Continuity and Differentiability
A polynomial p(x)p(x) when divided by x2−x^{2}- 3x+23 x+2 leaves remainder 2x−3.2 x-3 . Then
Q3MathsUnit 7: Limit, Continuity and Differentiability
\operatorname{Let} f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0<x \leq 2\end{array} and \right. g(x)=∣f(x)∣+f∣x∣\boldsymbol{g}(\boldsymbol{x})=|\boldsymbol{f}(\boldsymbol{x})|+\boldsymbol{f}|\boldsymbol{x}| then the number of points which g(x)g(x) is non differentiable, is
Q4MathsUnit 7: Limit, Continuity and Differentiability
Arrange the following limits in the ascending order: (1) lim⁡x→∞(1+x2+x)x+2\lim _{x \rightarrow \infty}\left(\frac{1+x}{2+x}\right)^{x+2} (2) lim⁡x→0(1+2x)3/x\lim _{x \rightarrow 0}(1+2 x)^{3 / x} (3) lim⁡θ→0sin⁡θ2θ\lim _{\boldsymbol{\theta} \rightarrow \mathbf{0}} \frac{\sin \boldsymbol{\theta}}{\mathbf{2} \boldsymbol{\theta}} (4) lim⁡x→0log⁡e(1+x)x\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{x}
Q5MathsUnit 7: Limit, Continuity and Differentiability
Assertion lim⁡x→01−cos⁡2xx\lim _{\boldsymbol{x} \rightarrow \mathbf{0}} \frac{\sqrt{1-\cos 2 x}}{\boldsymbol{x}} does not exist. Reason ∣sin⁡x∣={sin⁡x;0<x<π2−sin⁡x;−π2<x<0|\sin x|=\left\{\begin{array}{cc}\sin x ; & 0<x<\frac{\pi}{2} \\ -\sin x ; & -\frac{\pi}{2}<x<0\end{array}\right.
Q6MathsUnit 7: Limit, Continuity and Differentiability
Let f(x)\boldsymbol{f}(\boldsymbol{x}) be defined in the interval [-2,2] such that f(x)=f(x)= {−1,−2≤x≤0x−1,0<x≤2 and g(x)=\left\{\begin{array}{ll}-1, & -2 \leq x \leq 0 \\ x-1, & 0<x \leq 2\end{array} \text { and } g(x)=\right. f(∣x∣)+∣f(x)∣\boldsymbol{f}(|\boldsymbol{x}|)+|\boldsymbol{f}(\boldsymbol{x})| Test the differentiablity of g(x)g(x) in (-2,2)
Q7MathsUnit 7: Limit, Continuity and Differentiability
The function f(x)=f(x)= \left\{\begin{array}{l}\frac{\cos 3 x-\cos 4 x}{x^{2}}, \text { for } x \neq 0 \\ \frac{7}{2}, \text { for } x=0\end{array} at \right. x=0\boldsymbol{x}=\mathbf{0} is
Q8MathsUnit 7: Limit, Continuity and Differentiability
Assertion If f(x)=0f(x)=0 has two distinct positive real roots then number of non- differentiable points of y=∣f(−∣x∣)∣\boldsymbol{y}=|\boldsymbol{f}(-|\boldsymbol{x}|)| is 1\mathbf{1} Reason Graph of y=f(∣x∣)\boldsymbol{y}=\boldsymbol{f}(|\boldsymbol{x}|) is symmetrical about y-axis

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