OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Exercise 18 (D)

Question 1. \(\lim_{x \to 0} \frac { \sqrt{1+x}-1 }{ x }\)
Answer: To find the limit of the given expression as \(x\) approaches 0, we first notice that direct substitution results in the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). To resolve this, we multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{1+x}+1 \).
\( \lim_{x \to 0} \frac { \sqrt{1+x}-1 }{ x } \)
\( = \lim_{x \to 0} \frac { \sqrt{1+x}-1 }{ x } \times \frac { \sqrt{1+x}+1 }{ \sqrt{1+x}+1 } \)
\( = \lim_{x \to 0} \frac { (\sqrt{1+x})^2 - 1^2 }{ x(\sqrt{1+x}+1) } \)
\( = \lim_{x \to 0} \frac { (1+x)-1 }{ x(\sqrt{1+x}+1) } \)
\( = \lim_{x \to 0} \frac { x }{ x(\sqrt{1+x}+1) } \)
Now, we can cancel out \(x\) from the numerator and the denominator, as \(x \neq 0\) when taking the limit.
\( = \lim_{x \to 0} \frac { 1 }{ \sqrt{1+x}+1 } \)
Finally, substitute \(x=0\) into the simplified expression:
\( = \frac { 1 }{ \sqrt{1+0}+1 } \)
\( = \frac { 1 }{ \sqrt{1}+1 } \)
\( = \frac { 1 }{ 1+1 } \)
\( = \frac { 1 }{ 2 } \)
This method of multiplying by the conjugate is very useful for limits involving square roots that result in an indeterminate form.
In simple words: When you try to put \(x=0\) into the problem, you get \(0/0\). To fix this, you multiply the top and bottom by `\((\sqrt{1+x}+1)\)`. This helps to get rid of the square root on top and allows you to cancel `\(x\)`. Then, you can easily put \(x=0\) to get the answer.

🎯 Exam Tip: When faced with limits involving square roots that lead to an indeterminate form, remember to rationalize either the numerator or the denominator using its conjugate. This is a crucial first step.

Question 2. \(\lim_{x \to a} \frac { \sqrt{x}-\sqrt{a} }{ x-a }\)
Answer: To find the limit of the given expression as \(x\) approaches \(a\), we first notice that direct substitution results in the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). There are two main ways to solve this: by factoring the denominator or by multiplying by the conjugate.
**Method 1: Factoring the Denominator**
We can rewrite the denominator \(x-a\) as a difference of squares, \( (\sqrt{x})^2 - (\sqrt{a})^2 \).
\( \lim_{x \to a} \frac { \sqrt{x}-\sqrt{a} }{ x-a } \)
\( = \lim_{x \to a} \frac { \sqrt{x}-\sqrt{a} }{ (\sqrt{x})^2 - (\sqrt{a})^2 } \)
Using the difference of squares formula, \( A^2 - B^2 = (A-B)(A+B) \):
\( = \lim_{x \to a} \frac { \sqrt{x}-\sqrt{a} }{ (\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a}) } \)
Now, we can cancel out \( (\sqrt{x}-\sqrt{a}) \) from the numerator and the denominator, as \( x \neq a \).
\( = \lim_{x \to a} \frac { 1 }{ \sqrt{x}+\sqrt{a} } \)
Substitute \(x=a\) into the simplified expression:
\( = \frac { 1 }{ \sqrt{a}+\sqrt{a} } \)
\( = \frac { 1 }{ 2\sqrt{a} } \)
**Method 2: Multiplying by the Conjugate**
Multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x}+\sqrt{a} \).
\( \lim_{x \to a} \frac { \sqrt{x}-\sqrt{a} }{ x-a } \)
\( = \lim_{x \to a} \frac { \sqrt{x}-\sqrt{a} }{ x-a } \times \frac { \sqrt{x}+\sqrt{a} }{ \sqrt{x}+\sqrt{a} } \)
\( = \lim_{x \to a} \frac { (\sqrt{x})^2 - (\sqrt{a})^2 }{ (x-a)(\sqrt{x}+\sqrt{a}) } \)
\( = \lim_{x \to a} \frac { x-a }{ (x-a)(\sqrt{x}+\sqrt{a}) } \)
Cancel out \( (x-a) \) from the numerator and the denominator:
\( = \lim_{x \to a} \frac { 1 }{ \sqrt{x}+\sqrt{a} } \)
Substitute \(x=a\):
\( = \frac { 1 }{ \sqrt{a}+\sqrt{a} } \)
\( = \frac { 1 }{ 2\sqrt{a} } \)
Both methods yield the same result. The difference of squares factorization is a powerful algebraic tool that often helps simplify expressions involving square roots.
In simple words: This problem also gives \(0/0\) when you put \(x=a\). You can either break down the bottom part `\(x-a\)` into `\((\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})\)` or multiply the top and bottom by `\((\sqrt{x}+\sqrt{a})\)`. Both ways help you cancel out the part that makes it \(0/0\). Then, you put \(x=a\) to get the final answer.

🎯 Exam Tip: For limits involving expressions like \( \frac{\sqrt{x}-\sqrt{a}}{x-a} \), always remember that \( x-a \) can be written as \( (\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a}) \), which is key to simplification.

Question 3. \(\lim_{x \to 4} \frac { 3-\sqrt{5+x} }{ x-4 }\)
Answer: To find the limit of the given expression as \(x\) approaches 4, we first observe that direct substitution leads to the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). To solve this, we will multiply the numerator and the denominator by the conjugate of the numerator, which is \( 3+\sqrt{5+x} \).
\( \lim_{x \to 4} \frac { 3-\sqrt{5+x} }{ x-4 } \)
\( = \lim_{x \to 4} \frac { 3-\sqrt{5+x} }{ x-4 } \times \frac { 3+\sqrt{5+x} }{ 3+\sqrt{5+x} } \)
\( = \lim_{x \to 4} \frac { 3^2 - (\sqrt{5+x})^2 }{ (x-4)(3+\sqrt{5+x}) } \)
\( = \lim_{x \to 4} \frac { 9 - (5+x) }{ (x-4)(3+\sqrt{5+x}) } \)
\( = \lim_{x \to 4} \frac { 9 - 5 - x }{ (x-4)(3+\sqrt{5+x}) } \)
\( = \lim_{x \to 4} \frac { 4 - x }{ (x-4)(3+\sqrt{5+x}) } \)
Now, we can rewrite \( (4-x) \) as \( -(x-4) \).
\( = \lim_{x \to 4} \frac { -(x-4) }{ (x-4)(3+\sqrt{5+x}) } \)
Cancel out \( (x-4) \) from the numerator and the denominator, as \( x \neq 4 \).
\( = \lim_{x \to 4} \frac { -1 }{ 3+\sqrt{5+x} } \)
Substitute \(x=4\) into the simplified expression:
\( = \frac { -1 }{ 3+\sqrt{5+4} } \)
\( = \frac { -1 }{ 3+\sqrt{9} } \)
\( = \frac { -1 }{ 3+3 } \)
\( = \frac { -1 }{ 6 } \)
Recognizing that \( (4-x) \) is the negative of \( (x-4) \) is a key step in simplifying this type of limit problem.
In simple words: When you put \(x=4\) into this problem, you get \(0/0\). To solve it, you multiply the top and bottom by `\( (3+\sqrt{5+x}) \)`. This removes the square root on top. Then, you change `\(4-x\)` to `\( -(x-4) \)` so you can cancel `\(x-4\)` from the top and bottom. Finally, put \(x=4\) into the simple form to get the answer.

🎯 Exam Tip: Be careful with signs. If the numerator simplifies to \( (a-x) \) and the denominator has \( (x-a) \), remember that \( (a-x) = -(x-a) \) to allow for cancellation.

Question 4. \(\lim_{x \to 0} \frac { \sqrt{x+2}-\sqrt{2} }{ x }\)
Answer: To find the limit of the given expression as \(x\) approaches 0, we first check for direct substitution. This yields \(\left(\frac{0}{0}\text{ form}\right)\), which is an indeterminate form. To proceed, we multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{x+2}+\sqrt{2} \).
\( \lim_{x \to 0} \frac { \sqrt{x+2}-\sqrt{2} }{ x } \)
\( = \lim_{x \to 0} \frac { \sqrt{x+2}-\sqrt{2} }{ x } \times \frac { \sqrt{x+2}+\sqrt{2} }{ \sqrt{x+2}+\sqrt{2} } \)
\( = \lim_{x \to 0} \frac { (\sqrt{x+2})^2 - (\sqrt{2})^2 }{ x(\sqrt{x+2}+\sqrt{2}) } \)
\( = \lim_{x \to 0} \frac { (x+2)-2 }{ x(\sqrt{x+2}+\sqrt{2}) } \)
\( = \lim_{x \to 0} \frac { x }{ x(\sqrt{x+2}+\sqrt{2}) } \)
Now, we can cancel out \(x\) from the numerator and the denominator, since \(x \neq 0\) in the limit process.
\( = \lim_{x \to 0} \frac { 1 }{ \sqrt{x+2}+\sqrt{2} } \)
Finally, substitute \(x=0\) into the simplified expression:
\( = \frac { 1 }{ \sqrt{0+2}+\sqrt{2} } \)
\( = \frac { 1 }{ \sqrt{2}+\sqrt{2} } \)
\( = \frac { 1 }{ 2\sqrt{2} } \)
Always look for ways to eliminate the indeterminate form, and multiplying by the conjugate is a standard technique when square roots are involved.
In simple words: Putting \(x=0\) into the problem gives \(0/0\). To solve this, you multiply the top and bottom by `\((\sqrt{x+2}+\sqrt{2})\)`. This helps remove the square roots from the top. After that, you can cancel `\(x\)` from the top and bottom. Then, put \(x=0\) to get the final answer.

🎯 Exam Tip: When simplifying, ensure that you correctly apply the difference of squares formula \( (A-B)(A+B) = A^2-B^2 \) to the numerator after multiplying by the conjugate.

Question 5. \(\lim_{x \to 0} \frac { x }{ \sqrt{1+x}-1 }\)
Answer: To find the limit of the given expression as \(x\) approaches 0, we first notice that direct substitution results in the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). To resolve this, we multiply the numerator and the denominator by the conjugate of the denominator, which is \( \sqrt{1+x}+1 \).
\( \lim_{x \to 0} \frac { x }{ \sqrt{1+x}-1 } \)
\( = \lim_{x \to 0} \frac { x }{ \sqrt{1+x}-1 } \times \frac { \sqrt{1+x}+1 }{ \sqrt{1+x}+1 } \)
\( = \lim_{x \to 0} \frac { x(\sqrt{1+x}+1) }{ (\sqrt{1+x})^2 - 1^2 } \)
\( = \lim_{x \to 0} \frac { x(\sqrt{1+x}+1) }{ (1+x)-1 } \)
\( = \lim_{x \to 0} \frac { x(\sqrt{1+x}+1) }{ x } \)
Now, we can cancel out \(x\) from the numerator and the denominator, as \(x \neq 0\) when taking the limit.
\( = \lim_{x \to 0} (\sqrt{1+x}+1) \)
Finally, substitute \(x=0\) into the simplified expression:
\( = \sqrt{1+0}+1 \)
\( = \sqrt{1}+1 \)
\( = 1+1 \)
\( = 2 \)
Rationalizing the denominator is a common algebraic technique used to simplify expressions involving square roots in limits.
In simple words: This problem also gives \(0/0\) if you put \(x=0\) directly. Here, you multiply the top and bottom by `\((\sqrt{1+x}+1)\)` to get rid of the square root on the *bottom*. After that, you can cancel `\(x\)` from the top and bottom, making the problem simple enough to put \(x=0\) and find the answer.

🎯 Exam Tip: Remember to rationalize the part (numerator or denominator) that contains the square root causing the indeterminate form. In this case, it was the denominator.

Question 6. \(\lim_{x \to 0} \frac { \sqrt{1+x}-\sqrt{1-x} }{ 2x }\)
Answer: To find the limit of the given expression as \(x\) approaches 0, we first observe that direct substitution results in the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). To resolve this, we multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{1+x}+\sqrt{1-x} \).
\( \lim_{x \to 0} \frac { \sqrt{1+x}-\sqrt{1-x} }{ 2x } \)
\( = \lim_{x \to 0} \frac { \sqrt{1+x}-\sqrt{1-x} }{ 2x } \times \frac { \sqrt{1+x}+\sqrt{1-x} }{ \sqrt{1+x}+\sqrt{1-x} } \)
\( = \lim_{x \to 0} \frac { (\sqrt{1+x})^2 - (\sqrt{1-x})^2 }{ 2x(\sqrt{1+x}+\sqrt{1-x}) } \)
\( = \lim_{x \to 0} \frac { (1+x)-(1-x) }{ 2x(\sqrt{1+x}+\sqrt{1-x}) } \)
\( = \lim_{x \to 0} \frac { 1+x-1+x }{ 2x(\sqrt{1+x}+\sqrt{1-x}) } \)
\( = \lim_{x \to 0} \frac { 2x }{ 2x(\sqrt{1+x}+\sqrt{1-x}) } \)
Now, we can cancel out \(2x\) from the numerator and the denominator, as \(x \neq 0\) when taking the limit.
\( = \lim_{x \to 0} \frac { 1 }{ \sqrt{1+x}+\sqrt{1-x} } \)
Finally, substitute \(x=0\) into the simplified expression:
\( = \frac { 1 }{ \sqrt{1+0}+\sqrt{1-0} } \)
\( = \frac { 1 }{ \sqrt{1}+\sqrt{1} } \)
\( = \frac { 1 }{ 1+1 } \)
\( = \frac { 1 }{ 2 } \)
Remember to distribute the negative sign carefully when simplifying terms like \( (1+x)-(1-x) \) in the numerator.
In simple words: This limit problem also gives \(0/0\) if you put \(x=0\). To fix this, multiply the top and bottom by `\((\sqrt{1+x}+\sqrt{1-x})\)`. This helps remove the square roots from the top. Be careful when you subtract `\((1-x)\)`. After simplifying, you can cancel `\(2x\)` from both top and bottom. Then, put \(x=0\) to find the answer.

🎯 Exam Tip: When simplifying expressions like \( (1+x)-(1-x) \), pay close attention to the distribution of the negative sign to avoid common errors.

Question 7. \(\lim_{x \to 1} \frac { \sqrt{x^2-1}+\sqrt{x-1} }{ \sqrt{x^2-1} }\)
Answer: To find the limit of the given expression as \(x\) approaches 1, we first notice that direct substitution results in the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). To resolve this, we will factor the terms involving square roots.
We know that \( x^2-1 = (x-1)(x+1) \), so \( \sqrt{x^2-1} = \sqrt{(x-1)(x+1)} = \sqrt{x-1}\sqrt{x+1} \).
\( \lim_{x \to 1} \frac { \sqrt{x^2-1}+\sqrt{x-1} }{ \sqrt{x^2-1} } \)
Substitute \( \sqrt{x^2-1} \) with \( \sqrt{x-1}\sqrt{x+1} \):
\( = \lim_{x \to 1} \frac { \sqrt{x-1}\sqrt{x+1}+\sqrt{x-1} }{ \sqrt{x-1}\sqrt{x+1} } \)
Factor out \( \sqrt{x-1} \) from the terms in the numerator:
\( = \lim_{x \to 1} \frac { \sqrt{x-1}(\sqrt{x+1}+1) }{ \sqrt{x-1}\sqrt{x+1} } \)
Now, we can cancel out \( \sqrt{x-1} \) from the numerator and the denominator, as \( x \neq 1 \).
\( = \lim_{x \to 1} \frac { \sqrt{x+1}+1 }{ \sqrt{x+1} } \)
Finally, substitute \(x=1\) into the simplified expression:
\( = \frac { \sqrt{1+1}+1 }{ \sqrt{1+1} } \)
\( = \frac { \sqrt{2}+1 }{ \sqrt{2} } \)
This can also be written by splitting the fraction:
\( = \frac { \sqrt{2} }{ \sqrt{2} } + \frac { 1 }{ \sqrt{2} } \)
\( = 1 + \frac { 1 }{ \sqrt{2} } \)
Always look for opportunities to factor terms, especially when dealing with differences of squares or common factors under square roots.
In simple words: If you put \(x=1\) in this problem, you get \(0/0\). To solve it, break down `\(\sqrt{x^2-1}\)` into `\(\sqrt{x-1}\sqrt{x+1}\)`. Then, you can take `\(\sqrt{x-1}\)` as a common factor on top and cancel it with the bottom. Once it's simpler, you can put \(x=1\) to find the limit.

🎯 Exam Tip: Factoring the term \( x^2-1 \) into \( (x-1)(x+1) \) is crucial here. Remember that \( \sqrt{AB} = \sqrt{A}\sqrt{B} \).

Question 8. \(\lim_{x \to 0} \frac { \sqrt{1+x+x^2}-1 }{ x }\)
Answer: To find the limit of the given expression as \(x\) approaches 0, we first notice that direct substitution results in the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). To resolve this, we will multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{1+x+x^2}+1 \).
\( \lim_{x \to 0} \frac { \sqrt{1+x+x^2}-1 }{ x } \)
\( = \lim_{x \to 0} \frac { \sqrt{1+x+x^2}-1 }{ x } \times \frac { \sqrt{1+x+x^2}+1 }{ \sqrt{1+x+x^2}+1 } \)
\( = \lim_{x \to 0} \frac { (\sqrt{1+x+x^2})^2 - 1^2 }{ x(\sqrt{1+x+x^2}+1) } \)
\( = \lim_{x \to 0} \frac { (1+x+x^2)-1 }{ x(\sqrt{1+x+x^2}+1) } \)
\( = \lim_{x \to 0} \frac { x+x^2 }{ x(\sqrt{1+x+x^2}+1) } \)
Factor out \(x\) from the terms in the numerator:
\( = \lim_{x \to 0} \frac { x(1+x) }{ x(\sqrt{1+x+x^2}+1) } \)
Now, we can cancel out \(x\) from the numerator and the denominator, as \(x \neq 0\) when taking the limit.
\( = \lim_{x \to 0} \frac { 1+x }{ \sqrt{1+x+x^2}+1 } \)
Finally, substitute \(x=0\) into the simplified expression:
\( = \frac { 1+0 }{ \sqrt{1+0+0^2}+1 } \)
\( = \frac { 1 }{ \sqrt{1}+1 } \)
\( = \frac { 1 }{ 1+1 } \)
\( = \frac { 1 }{ 2 } \)
Remember that \( x^2 \) also approaches 0 as \( x \) approaches 0, which simplifies the square root term quickly.
In simple words: This problem gives \(0/0\) when you put \(x=0\). To solve it, multiply the top and bottom by `\((\sqrt{1+x+x^2}+1)\)`. This clears the square root from the top. Then, you can take `\(x\)` as a common factor on top and cancel it with the `\(x\)` on the bottom. Finally, put \(x=0\) into the simpler form to get the answer.

🎯 Exam Tip: Always factor out the common term (like `\(x\)` in this case) from the numerator after rationalizing, as it's often the factor causing the indeterminate form.

Question 9. \(\lim_{x \to 0} \frac { \sqrt{1+x^3}-\sqrt{1-x^3} }{ x^2 }\)
Answer: To find the limit of the given expression as \(x\) approaches 0, we first observe that direct substitution results in the indeterminate form \(\left(\frac{0}{0}\text{ form}\right)\). To resolve this, we will multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{1+x^3}+\sqrt{1-x^3} \).
\( \lim_{x \to 0} \frac { \sqrt{1+x^3}-\sqrt{1-x^3} }{ x^2 } \)
\( = \lim_{x \to 0} \frac { \sqrt{1+x^3}-\sqrt{1-x^3} }{ x^2 } \times \frac { \sqrt{1+x^3}+\sqrt{1-x^3} }{ \sqrt{1+x^3}+\sqrt{1-x^3} } \)
\( = \lim_{x \to 0} \frac { (\sqrt{1+x^3})^2 - (\sqrt{1-x^3})^2 }{ x^2(\sqrt{1+x^3}+\sqrt{1-x^3}) } \)
\( = \lim_{x \to 0} \frac { (1+x^3)-(1-x^3) }{ x^2(\sqrt{1+x^3}+\sqrt{1-x^3}) } \)
\( = \lim_{x \to 0} \frac { 1+x^3-1+x^3 }{ x^2(\sqrt{1+x^3}+\sqrt{1-x^3}) } \)
\( = \lim_{x \to 0} \frac { 2x^3 }{ x^2(\sqrt{1+x^3}+\sqrt{1-x^3}) } \)
Now, we can cancel out \(x^2\) from the numerator and the denominator, as \(x \neq 0\) when taking the limit.
\( = \lim_{x \to 0} \frac { 2x }{ \sqrt{1+x^3}+\sqrt{1-x^3} } \)
Finally, substitute \(x=0\) into the simplified expression:
\( = \frac { 2 \times 0 }{ \sqrt{1+0^3}+\sqrt{1-0^3} } \)
\( = \frac { 0 }{ \sqrt{1}+\sqrt{1} } \)
\( = \frac { 0 }{ 1+1 } \)
\( = \frac { 0 }{ 2 } \)
\( = 0 \)
Be careful with the minus sign when simplifying \( (1+x^3)-(1-x^3) \), as it needs to be distributed to both terms inside the second parenthesis.
In simple words: This limit problem gives \(0/0\) when you put \(x=0\). To fix this, multiply the top and bottom by `\((\sqrt{1+x^3}+\sqrt{1-x^3})\)`. This removes the square roots from the top. After that, you can cancel `\(x^2\)` from both top and bottom. Then, put \(x=0\) to find the final answer, which in this case is 0.

🎯 Exam Tip: Pay close attention to the algebraic simplification steps, especially when expanding \( (A-B)(A+B) \) and dealing with negative signs in the numerator, as a small error can lead to an incorrect result.