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 method since the function $f(x)\; =\; x+3$ is a polynomial and is continuous everywhere. We substitute $x\; =\; 3$ directly into the expression:

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

Thus, the value of the limit is 6.

The given limit is $\backslash lim\_\{x\; \backslash to\; \backslash pi\}\; \backslash left(\; x\; -\; \backslash frac\{22\}\{7\}\; \backslash right)$. This is a limit of a linear function. We can evaluate this limit by direct substitution of $x\; =\; \backslash pi$ into the expression, as the function is continuous.

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

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

We need to evaluate the limit $\backslash lim\_\{r\backslash to\; 1\}\; \backslash pi\; r^2$. Since $\backslash pi\; r^2$ is a polynomial in $r$, it is continuous. We can find the limit by substituting $r\; =\; 1$ directly into the expression.

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

The value of the limit is $\backslash pi$.

To evaluate the limit $\backslash lim\_\{x\backslash to\; 4\}\; \backslash frac\{4x+3\}\{x-2\}$, we first check if direct substitution is possible. The denominator $x-2$ is not zero when $x=4$. Therefore, we can substitute $x=4$ directly into the expression:

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

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

We need to evaluate the limit $\backslash lim\_\{x\; \backslash to\; -1\}\; \backslash frac\{x^\{10\}\; +\; x^5\; +\; 1\}\{x\; -\; 1\}$. The function is a rational function. We can attempt direct substitution of $x\; =\; -1$ into the expression, as the denominator will not be zero.

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

Now, we simplify the terms:

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

The value of the limit is $-\; \backslash frac\{1\}\{2\}$.

We are asked to evaluate the limit $\backslash lim\_\{x\backslash to\; 0\}\; \backslash frac\{(x+1)^5-1\}\{x\}$. If we substitute $x=0$ directly, 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 resolve this, we can use a substitution or the standard limit formula for $\backslash frac\{x^n\; -\; a^n\}\{x\; -\; a\}$.

Method 1: Using substitution

Let $y\; =\; x+1$. 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 is now in the form $\backslash lim\_\{y\backslash to\; 1\}\; \backslash frac\{y^5\; -\; 1^5\}\{y\; -\; 1\}$. We can use the standard limit formula $\backslash lim\_\{x\; \backslash to\; a\}\; \backslash frac\{x^n\; -\; a^n\}\{x\; -\; a\}\; =\; na^\{n-1\}$.

Here, $n=5$ and $a=1$.

$$=\; 5(1)^\{5-1\}\; =\; 5(1)^4\; =\; 5\; \backslash times\; 1\; =\; 5$$

Method 2: Using binomial expansion (alternative approach)

We can expand $(x+1)^5$ using the binomial theorem: $(x+1)^5\; =\; \backslash binom\{5\}\{0\}x^5\; +\; \backslash binom\{5\}\{1\}x^4\; +\; \backslash binom\{5\}\{2\}x^3\; +\; \backslash binom\{5\}\{3\}x^2\; +\; \backslash binom\{5\}\{4\}x^1\; +\; \backslash binom\{5\}\{5\}x^0$

$(x+1)^5\; =\; 1\; +\; 5x\; +\; 10x^2\; +\; 10x^3\; +\; 5x^4\; +\; x^5$

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

Now, divide by $x$ (for $x\; \backslash neq\; 0$):

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

Now, take the limit as $x\; \backslash to\; 0$:

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

Both methods yield the same result. The value of the limit is 5.

We need to evaluate the limit $\backslash lim\_\{x\; \backslash to\; 2\}\; \backslash frac\{3x^2\; -\; x\; -\; 10\}\{x^2\; -\; 4\}$. If we substitute $x=2$ directly, the numerator becomes $3(2)^2\; -\; 2\; -\; 10\; =\; 3(4)\; -\; 2\; -\; 10\; =\; 12\; -\; 12\; =\; 0$, and the denominator becomes $2^2\; -\; 4\; =\; 4\; -\; 4\; =\; 0$. This results in the indeterminate form $\backslash frac\{0\}\{0\}$.

To resolve this, we need to factorize the numerator and the denominator to cancel out the common factor $(x-2)$.

Factorizing the numerator $3x^2\; -\; x\; -\; 10$:

We look for two numbers that multiply to $3\; \backslash times\; -10\; =\; -30$ and add up to $-1$. These numbers are $-6$ and $5$.

$3x^2\; -\; x\; -\; 10\; =\; 3x^2\; -\; 6x\; +\; 5x\; -\; 10\; =\; 3x(x\; -\; 2)\; +\; 5(x\; -\; 2)\; =\; (3x\; +\; 5)(x\; -\; 2)$

Factorizing the denominator $x^2\; -\; 4$:

This is a difference of squares: $x^2\; -\; 4\; =\; (x\; -\; 2)(x\; +\; 2)$.

Now, substitute the factorized forms back into the limit expression:

$$\backslash lim\_\{x\; \backslash to\; 2\}\; \backslash frac\{(3x\; +\; 5)(x\; -\; 2)\}\{(x\; -\; 2)(x\; +\; 2)\}$$

Since $x\; \backslash to\; 2$, $x\; \backslash neq\; 2$, so we can cancel the $(x\; -\; 2)$ term:

$$=\; \backslash lim\_\{x\backslash to\; 2\}\; \backslash frac\{3x+5\}\{x+2\}$$

Now, we can use direct substitution again:

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

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