4. (a) Evaluate: Lt (1/x^2) tan x as x approaches 0. (b) Calculate the approximate value of √26 to three decimal places by Taylor's expansion. 5. (a) State and prove Lagrange's mea... 4. (a) Evaluate: Lt (1/x^2) tan x as x approaches 0. (b) Calculate the approximate value of √26 to three decimal places by Taylor's expansion. 5. (a) State and prove Lagrange's mean value theorem. (b) Verify Rolle's theorem for the function e^x sin x in [0, π]. 6. (a) Examine the curve y = x^4 - 2x^3 + 1 for concavity upwards, concavity downwards and points of inflexion. Read more

Steps to Solve

1. Limit Calculation Setup
   We need to evaluate the limit:
   $$ \lim_{x \to 0} \left(\frac{1}{x^2}\right) \tan x $$

2. Substituting Taylor Series for $\tan x$
   Using the Taylor series expansion for $\tan x$ around $x = 0$:
   $$ \tan x = x + \frac{x^3}{3} + O(x^5) $$

3. Plug in the Series into the Limit
   Substituting the Taylor series into the limit expression:
   $$ \lim_{x \to 0} \left(\frac{1}{x^2}\right) \left(x + \frac{x^3}{3} + O(x^5)\right) $$

4. Simplifying the Expression
   This gives:
   $$ \lim_{x \to 0} \left( \frac{1}{x^2} \cdot x + \frac{1}{x^2} \cdot \frac{x^3}{3} + \frac{1}{x^2} \cdot O(x^5) \right) $$
   Which simplifies to:
   $$ \lim_{x \to 0} \left( \frac{1}{x} + \frac{x}{3} + O(x^3) \right) $$

5. Evaluating the Limit
   As $x \to 0$:
   The first term $\frac{1}{x}$ goes to $\infty$. Thus, the limit diverges.

6. Approximate Value of $\sqrt{26}$ using Taylor Series
   Next, expanding the function $\sqrt{x}$ around $x=25$:
   Using the expansion:
   $$ \sqrt{x} \approx 5 + \frac{1}{2\sqrt{25}}(x - 25) $$

7. Plug in $x = 26$
   Now, substituting $x = 26$:
   $$ \sqrt{26} \approx 5 + \frac{1}{10}(26 - 25) = 5 + 0.1 = 5.1 $$

8. Rounding Off to Three Decimal Places
   Thus, the approximate value of $\sqrt{26}$ to three decimal places is:
   $$ \sqrt{26} \approx 5.100 $$