JEE MATHS PRACTICE - MAINS TO ADVANCED - CALCULUS - LIMITS, CONTINUITY AND DIFFERENTIABILITY

Here are 25 concept-based solved examples on Limits, Continuity, and Differentiability with increasing difficulty:

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PART 1: Basic Problems (Easy Level)

1. Evaluate $\lim\limits_{x \to 2} (x^2 + 3x - 4)$
Solution: Direct substitution gives $2^2 + 3(2) - 4 = 4 + 6 - 4 = 6$.

2. Find $\lim\limits_{x \to 0} \frac{\sin x}{x}$.
Solution: Standard result: $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

3. Evaluate $\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2}$.
Solution: Using $\cos x \approx 1 - \frac{x^2}{2}$ for small $x$, we get:

$$\frac{1 - (1 - x^2/2)}{x^2} = \frac{x^2/2}{x^2} = \frac{1}{2}$$

4. Check continuity of $f(x) = \begin{cases} x^2, & x \leq 1 \\ 2x - 1, & x > 1 \end{cases}$ at $x = 1$.
Solution: $\lim\limits_{x \to 1^-} f(x) = 1^2 = 1$, $\lim\limits_{x \to 1^+} f(x) = 2(1) - 1 = 1$, and $f(1) = 1$. So, $f(x)$ is continuous at $x = 1$.

5. Evaluate $\lim\limits_{x \to \infty} \frac{3x^2 + 2}{5x^2 + 7}$.
Solution: Divide by $x^2$:

$$\lim\limits_{x \to \infty} \frac{3 + 2/x^2}{5 + 7/x^2} = \frac{3}{5}$$

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PART 2: Intermediate Level Problems

6. Evaluate $\lim\limits_{x \to 0} \frac{x - \sin x}{x^3}$.
Solution: Using Maclaurin series:

$$\sin x \approx x - \frac{x^3}{6} \Rightarrow x - \sin x \approx \frac{x^3}{6}$$ $$\lim\limits_{x \to 0} \frac{x - \sin x}{x^3} = \frac{\frac{x^3}{6}}{x^3} = \frac{1}{6}$$

7. Find the derivative of $f(x) = x^x$.
Solution: Take logarithm:

$$y = x^x \Rightarrow \ln y = x \ln x$$

Differentiate:

$$\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1$$ $$f'(x) = x^x (\ln x + 1)$$

8. Show that $f(x) = x^3 - 3x + 2$ is differentiable at $x = 1$.
Solution:

$$f'(x) = 3x^2 - 3$$ $$f'(1) = 3(1)^2 - 3 = 0$$

9. Check differentiability of $f(x) = |x|$ at $x = 0$.
Solution: Left derivative: $f'_-(0) = -1$, right derivative: $f'_+(0) = 1$. Since they are unequal, $f(x)$ is not differentiable at $x = 0$.

10. Evaluate $\lim\limits_{x \to 0} \frac{\tan x - x}{x^3}$.
Solution: Using Maclaurin expansion:

$$\tan x = x + \frac{x^3}{3} + O(x^5)$$ $$\lim\limits_{x \to 0} \frac{\tan x - x}{x^3} = \frac{x + x^3/3 - x}{x^3} = \frac{x^3/3}{x^3} = \frac{1}{3}$$

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PART 3: Advanced & Challenging Problems

11. Prove that $f(x) = x^3 + x^2 + x + 1$ is continuous everywhere.
Solution: Polynomial functions are always continuous.

12. Evaluate $\lim\limits_{x \to 0} \frac{e^x - 1 - x}{x^2}$.
Solution: Using Maclaurin series,

$$e^x = 1 + x + x^2/2 + O(x^3)$$ $$\lim\limits_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim\limits_{x \to 0} \frac{x^2/2 + O(x^3)}{x^2} = \frac{1}{2}$$

13. If $f(x) = x^3 + 3x^2 + 5x + 7$, find points where $f'(x) = 0$.
Solution:

$$f'(x) = 3x^2 + 6x + 5$$

Solving $3x^2 + 6x + 5 = 0$, we get no real roots.

14. Prove that $g(x) = x^3 - 6x^2 + 11x - 6$ is differentiable everywhere.
Solution: Since it is a polynomial, it is differentiable everywhere.

15. Evaluate $\lim\limits_{x \to 0} \frac{\ln(1+x)}{x}$.
Solution: Using Maclaurin series,

$$\ln(1+x) = x - x^2/2 + O(x^3)$$ $$\lim\limits_{x \to 0} \frac{x - x^2/2}{x} = 1$$

16-25. Challenge Problems:
These will involve epsilon-delta proofs, L'Hôpital's Rule, Rolle’s Theorem, Mean Value Theorem, and function differentiability proofs. 

Here are 10 challenging problems (16-25) on Limits, Continuity, and Differentiability, including advanced proofs and theorem applications.

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PART 4: Challenging Problems (Advanced Level)

16. Prove that $\lim\limits_{x \to 0} \frac{e^x - 1}{x} = 1$ using the epsilon-delta definition.

Solution:
For any $\epsilon > 0$, we must find $\delta > 0$ such that

$$0 < |x| < \delta \Rightarrow \left| \frac{e^x - 1}{x} - 1 \right| < \epsilon.$$

Using Taylor expansion:

$$e^x = 1 + x + \frac{x^2}{2} + O(x^3).$$ $$\frac{e^x - 1}{x} = 1 + \frac{x}{2} + O(x^2).$$

Choosing $\delta = \min(1, \frac{2\epsilon}{3})$, we ensure $\left| \frac{e^x - 1}{x} - 1 \right| < \epsilon$.

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17. Use L'Hôpital’s Rule to evaluate $\lim\limits_{x \to 0} \frac{x^2}{e^x - 1 - x}$.

Solution:
Direct substitution gives $\frac{0}{0}$. Differentiating numerator and denominator:

$$\text{Numerator: } \frac{d}{dx} (x^2) = 2x.$$ $$\text{Denominator: } \frac{d}{dx} (e^x - 1 - x) = e^x - 1.$$

Applying L'Hôpital’s Rule:

$$\lim\limits_{x \to 0} \frac{2x}{e^x - 1} = \lim\limits_{x \to 0} \frac{2x}{x + x^2/2} = 2.$$

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18. Use Rolle’s Theorem to prove that $f(x) = x^3 - 3x + 1$ has at least one root in $(-2,2)$.

Solution:
Since $f(x)$ is polynomial, it is continuous and differentiable everywhere.

$$f(-2) = (-8 + 6 + 1) = -1, \quad f(2) = (8 - 6 + 1) = 3.$$

By Intermediate Value Theorem, there exists $c \in (-2,2)$ such that $f(c) = 0$.

Now, Rolle’s Theorem states $f'(c) = 0$ for some $c$:

$$f'(x) = 3x^2 - 3, \quad \text{solving } 3x^2 - 3 = 0 \Rightarrow x = \pm 1.$$

Since $x = 1$ and $x = -1$ are in $(-2,2)$, Rolle’s Theorem is verified.

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19. Verify Mean Value Theorem for $f(x) = x^3 - 6x$ in $[-2,2]$.

Solution:
Since $f(x)$ is continuous and differentiable on $(-2,2)$, MVT applies:

$$f'(c) = \frac{f(2) - f(-2)}{2 - (-2)}.$$ $$f(2) = (8 - 12) = -4, \quad f(-2) = (-8 + 12) = 4.$$ $$\frac{-4 - 4}{4} = -2.$$ $$f'(x) = 3x^2 - 6, \quad \text{solving } 3c^2 - 6 = -2 \Rightarrow c^2 = \frac{4}{3}.$$

Thus, $c = \pm \frac{2}{\sqrt{3}}$ exists in $(-2,2)$.

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20. Show that $f(x) = |x - 2|^3$ is continuous but not differentiable at $x = 2$.

Solution:

• Continuity:

$$\lim\limits_{x \to 2} f(x) = |2-2|^3 = 0, \quad f(2) = 0.$$

So, $f(x)$ is continuous.

• Differentiability:

$$\lim\limits_{h \to 0^+} \frac{(2+h-2)^3 - 0}{h} = \lim\limits_{h \to 0^+} \frac{h^3}{h} = 0.$$ $$\lim\limits_{h \to 0^-} \frac{(2-h-2)^3 - 0}{h} = \lim\limits_{h \to 0^-} \frac{(-h)^3}{h} = -h^2.$$

Since left derivative $= 0$ and right derivative $= 0$, function is not differentiable.

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21. Find $\lim\limits_{x \to 0} \frac{x^3}{\sin^3 x}$.

Solution:
Using $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$, we get

$$\lim\limits_{x \to 0} \frac{x^3}{\sin^3 x} = 1.$$

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22. Prove that $f(x) = x^3 - 5x + 2$ has a root in $(0,2)$.

Solution:

$$f(0) = 2, \quad f(2) = 8 - 10 + 2 = 0.$$

By Intermediate Value Theorem, $f(x)$ has at least one root in $(0,2)$.

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23. Evaluate $\lim\limits_{x \to 0} \frac{x - \tan x}{x^3}$.

Solution:

$$\tan x = x + \frac{x^3}{3} + O(x^5),$$ $$\lim\limits_{x \to 0} \frac{x - (x + x^3/3)}{x^3} = \lim\limits_{x \to 0} \frac{-x^3/3}{x^3} = -\frac{1}{3}.$$

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24. Find the derivative of $f(x) = x^{x^x}$.

Solution:
Taking natural log:

$$\ln y = x^x \ln x.$$

Differentiating:

$$\frac{1}{y} \cdot \frac{dy}{dx} = x^x (\ln x + 1) + x^x \frac{1}{x}.$$ $$f'(x) = x^{x^x} \left( x^x (\ln x + 1) + x^{x-1} \right).$$

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25. Show that $f(x) = |x^2 - 4|$ is not differentiable at $x = \pm2$.

Solution:

$$f'(x) = 2x, \quad \text{but at } x = 2, f'(2^-) \neq f'(2^+).$$

Thus, $f(x)$ is not differentiable at $x = \pm 2$.

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This completes 25 solved examples on Limits, Continuity, and Differentiability with increasing difficulty.