CBSE Class 11 Mathematics Chapter 13: Limits and Derivatives NCERT Solutions

To evaluate the limit $\backslash lim\_\{x\backslash to\; 3\}\; (x+3)$, we can use the direct substitution property of limits for polynomial functions. Since $f(x)\; =\; x+3$ is a polynomial, it is continuous everywhere. Therefore, we can find the limit by substituting $x=3$ directly into the function.

$$\backslash lim\_\{x\backslash to\; 3\}\; (x+3)\; =\; 3\; +\; 3\; =\; 6$$

The limit of the function as $x$ approaches 3 is 6.

The function $f(x)\; =\; x\; -\; \backslash frac\{22\}\{7\}$ is a polynomial in $x$, which is continuous everywhere. We can evaluate the limit by direct substitution of $x\; =\; \backslash pi$.

$$\backslash lim\_\{x\; \backslash to\; \backslash pi\}\; \backslash left(\; x\; -\; \backslash frac\{22\}\{7\}\; \backslash right)\; =\; \backslash pi\; -\; \backslash frac\{22\}\{7\}$$

The value of the limit is $\backslash pi\; -\; \backslash frac\{22\}\{7\}$.

The function $f(r)\; =\; \backslash pi\; r^2$ is a polynomial in $r$ multiplied by a constant $\backslash pi$. It is continuous everywhere. We can find the limit by substituting $r=1$ directly into the function.

$$\backslash lim\_\{r\backslash to\; 1\}\; \backslash pi\; r^2\; =\; \backslash pi\; (1)^2\; =\; \backslash pi\; \backslash times\; 1\; =\; \backslash pi$$

The limit of the function as $r$ approaches 1 is $\backslash pi$.

The given function is a rational function $f(x)\; =\; \backslash frac\{4x+3\}\{x-2\}$. We first check if direct substitution of $x=4$ is possible. The denominator is $4-2\; =\; 2$, which is not zero. Thus, we can use direct substitution.

$$\backslash lim\_\{x\; \backslash to\; 4\}\; \backslash frac\{4x+3\}\{x-2\}\; =\; \backslash frac\{4(4)+3\}\{4-2\}\; =\; \backslash frac\{16+3\}\{2\}\; =\; \backslash frac\{19\}\{2\}$$

The value of the limit is $\backslash frac\{19\}\{2\}$.

The function is a rational function $f(x)\; =\; \backslash frac\{x^\{10\}\; +\; x^5\; +\; 1\}\{x\; -\; 1\}$. We check the denominator at $x\; =\; -1$. The denominator is $-1\; -\; 1\; =\; -2$, which is not zero. Therefore, we can evaluate the limit using direct substitution.

$$\backslash lim\_\{x\; \backslash to\; -1\}\; \backslash frac\{x^\{10\}\; +\; x^5\; +\; 1\}\{x\; -\; 1\}\; =\; \backslash frac\{(-1)^\{10\}\; +\; (-1)^5\; +\; 1\}\{-1\; -\; 1\}$$

Calculating the powers of -1: $(-1)^\{10\}\; =\; 1$ and $(-1)^5\; =\; -1$.

$$=\; \backslash frac\{1\; +\; (-1)\; +\; 1\}\{-2\}\; =\; \backslash frac\{1\; -\; 1\; +\; 1\}\{-2\}\; =\; \backslash frac\{1\}\{-2\}\; =\; -\backslash frac\{1\}\{2\}$$

The limit of the function as $x$ approaches -1 is $-1/2$.

When we try to substitute $x=0$ directly into the expression $\backslash frac\{(x+1)^5-1\}\{x\}$, we get $\backslash frac\{(0+1)^5-1\}\{0\}\; =\; \backslash frac\{1^5-1\}\{0\}\; =\; \backslash frac\{0\}\{0\}$, which is an indeterminate form. To evaluate this limit, we can use a substitution or recognize it as a form related to the definition of the derivative.

Let's use the substitution $y\; =\; x+1$. As $x\; \backslash to\; 0$, we have $y\; \backslash to\; 0+1$, so $y\; \backslash to\; 1$. Also, $x\; =\; y-1$.

Substituting these into the limit expression:

$$\backslash lim\_\{x\backslash to\; 0\}\; \backslash frac\{(x+1)^5\; -\; 1\}\{x\}\; =\; \backslash lim\_\{y\backslash to\; 1\}\; \backslash frac\{y^5\; -\; 1\}\{y-1\}$$

This expression is in the form $\backslash lim\_\{y\backslash to\; a\}\; \backslash frac\{y^n\; -\; a^n\}\{y\; -\; a\}$, where $n=5$ and $a=1$. The value of this limit is given by the formula $na^\{n-1\}$.

$$=\; 5\; \backslash cdot\; 1^\{5-1\}\; =\; 5\; \backslash cdot\; 1^4\; =\; 5\; \backslash cdot\; 1\; =\; 5$$

Alternatively, we can expand $(x+1)^5$ using the binomial theorem: $(x+1)^5\; =\; x^5\; +\; 5x^4\; +\; 10x^3\; +\; 10x^2\; +\; 5x\; +\; 1$.

So, $(x+1)^5\; -\; 1\; =\; x^5\; +\; 5x^4\; +\; 10x^3\; +\; 10x^2\; +\; 5x$.

Then, $\backslash frac\{(x+1)^5\; -\; 1\}\{x\}\; =\; \backslash frac\{x^5\; +\; 5x^4\; +\; 10x^3\; +\; 10x^2\; +\; 5x\}\{x\}$.

For $x\; \backslash neq\; 0$, we can divide each term by $x$: $x^4\; +\; 5x^3\; +\; 10x^2\; +\; 10x\; +\; 5$.

Now, we can evaluate the limit by direct substitution:

$$\backslash lim\_\{x\backslash to\; 0\}\; (x^4\; +\; 5x^3\; +\; 10x^2\; +\; 10x\; +\; 5)\; =\; 0^4\; +\; 5(0)^3\; +\; 10(0)^2\; +\; 10(0)\; +\; 5\; =\; 5$$

Thus, the limit is 5.

First, we attempt direct substitution of x=2 into the expression \frac{3x^2 - x - 10}{x^2 - 4}.

Numerator: 3(2)^2 - 2 - 10 = 3(4) - 2 - 10 = 12 - 2 - 10 = 0.

Denominator: (2)^2 - 4 = 4 - 4 = 0.

Since we get the indeterminate form \frac{0}{0}, we need to simplify the expression algebraically.

We factor the numerator and the denominator.

The denominator is a difference of squares: x^2 - 4 = (x-2)(x+2).

For the numerator 3x^2 - x - 10, we look for two numbers that multiply to 3 \times -10 = -30 and add up to -1. These numbers are -6 and 5.

So, 3x^2 - x - 10 = 3x^2 - 6x + 5x - 10 = 3x(x-2) + 5(x-2) = (3x+5)(x-2).

Now, substitute the factored forms back into the limit:

\lim_{x \to 2} \frac{(3x+5)(x-2)}{(x-2)(x+2)}

Since x \to 2, x \neq 2, so x-2 \neq 0. We can cancel the (x-2) term from the numerator and denominator:

= \lim_{x\to 2} \frac{3x+5}{x+2}

Now, we can use direct substitution again:

= \frac{3(2)+5}{2+2} = \frac{6+5}{4} = \frac{11}{4}

The value of the limit is \frac{11}{4}.