Evaluate the following limits: 

\displaystyle \text{Question 1: } \lim \limits_{x \to \infty} \frac{(3x-1)(4x-2)}{(x+8)(x-1)}

Answer:

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 2. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x^2.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0  

\displaystyle  \lim \limits_{x \to \infty} \frac{(3x-1)(4x-2)}{(x+8)(x-1)}

\displaystyle  = \lim \limits_{x \to \infty} \frac{ \Big(\frac{3x-1}{x}\Big)  \Big( \frac{4x-2}{x} \Big)  }{  \Big( \frac{x+8}{x}  \Big)  \Big( \frac{x-1}{x} \Big)} = \lim \limits_{x \to \infty} \frac{ \Big(3 - \frac{1}{x}\Big)  \Big( 4 - \frac{2}{x} \Big)  }{  \Big( 1+ \frac{8}{x}  \Big)  \Big( 1 - \frac{1}{x} \Big)} = \frac{3 \times 4}{1} = 1

\\

\displaystyle \text{Question 2: } \lim \limits_{x \to \infty} \frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}

Answer:

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 3. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x^3.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x}, \frac{1}{x^2}, \frac{1}{x^3} \rightarrow 0

\displaystyle \lim \limits_{x \to \infty} \frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7} = \lim \limits_{x \to \infty} \frac{3-\frac{4}{x}+\frac{6}{x^2}-1}{2+\frac{1}{x}-\frac{5}{x^2}+\frac{7}{x^3}} = \frac{3}{2}

\\

\displaystyle \text{Question 3: } \lim \limits_{x \to \infty} \frac{5x^3-6}{\sqrt{9+4x^6}}

Answer:

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 3. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x^3.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x^3}, \frac{1}{x^6} \rightarrow 0

\displaystyle \lim \limits_{x \to \infty} \frac{5x^3-6}{\sqrt{9+4x^6}} = \lim \limits_{x \to \infty} \frac{\frac{5x^3-6}{x^3}}{\frac{\sqrt{9+4x^6}}{x^3}} = \lim \limits_{x \to \infty} \frac{5 - \frac{6}{x^3}}{\sqrt{\frac{9}{x^6}+4}} = \frac{5}{2}

\\

\displaystyle \text{Question 4: } \lim \limits_{x \to \infty} \sqrt{x^2+cx} -x

Answer:

\displaystyle  \lim \limits_{x \to \infty} \sqrt{x^2+cx} -x

This is of the form \displaystyle  \infty - \infty.

Rationalizing the numerator:

\displaystyle  = \lim \limits_{x \to \infty} (\sqrt{x^2+cx} -x) \times \frac{\sqrt{x^2+cx} +x}{\sqrt{x^2+cx} +x}

\displaystyle  = \lim \limits_{x \to \infty} \ \frac{x^2+cx - x^2}{\sqrt{x^2+cx} +x}

\displaystyle  = \lim \limits_{x \to \infty} \ \frac{cx }{\sqrt{x^2+cx} +x}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0  

\displaystyle  = \lim \limits_{x \to \infty} \ \frac{c}{\sqrt{1+\frac{c}{x}} +1} = \frac{c}{\sqrt{1}+1} = \frac{c}{2}

\\

\displaystyle \text{Question 5: } \lim \limits_{x \to \infty} \sqrt{x+1}-\sqrt{x}

Answer:

\displaystyle \lim \limits_{x \to \infty} \sqrt{x+1}-\sqrt{x}

This is of the form \displaystyle  \infty - \infty.

Rationalizing the numerator:

\displaystyle = \lim \limits_{x \to \infty} (\sqrt{x+1}-\sqrt{x}) \times \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}

\displaystyle = \lim \limits_{x \to \infty} \frac{x+1 - x }{\sqrt{x+1}+\sqrt{x}}

\displaystyle = \lim \limits_{x \to \infty} \frac{1 }{\sqrt{x+1}+\sqrt{x}}

\displaystyle = \frac{1}{\infty} = 0

\\

\displaystyle \text{Question 6: } \lim \limits_{x \to \infty} \sqrt{x^2+7x} - x

Answer:

\displaystyle \lim \limits_{x \to \infty} \sqrt{x^2+7x} - x

This is of the form \displaystyle  \infty - \infty.

Rationalizing the numerator:

\displaystyle = \lim \limits_{x \to \infty} (\sqrt{x^2+7x} - x) \times \frac{\sqrt{x^2+7x} + x}{\sqrt{x^2+7x} + x}

\displaystyle = \lim \limits_{x \to \infty} \frac{x^2 + 7x - x^2}{\sqrt{x^2+7x} + x}

\displaystyle = \lim \limits_{x \to \infty} \frac{7x}{\sqrt{x^2+7x} + x}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0  

\displaystyle = \lim \limits_{x \to \infty} \frac{7}{\sqrt{1+\frac{7}{x}} + 1}

\displaystyle = \frac{7}{\sqrt{1}+1} = \frac{7}{2}

\\

\displaystyle \text{Question 7: } \lim \limits_{x \to \infty} \frac{x}{\sqrt{4x^2+1}-1}

Answer:

\displaystyle  \lim \limits_{x \to \infty} \frac{x}{\sqrt{4x^2+1}-1}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x}, \frac{1}{x^2} \rightarrow 0

\displaystyle  = \lim \limits_{x \to \infty} \frac{1}{\sqrt{4+\frac{1}{x^2}}-\frac{1}{x}}

\displaystyle  = \frac{1}{\sqrt{4}} = \frac{1}{2}

\\

\displaystyle \text{Question 8: } \lim \limits_{n \to \infty} \frac{n^2}{1+2+3+\cdots + n}

Answer:

\displaystyle \lim \limits_{n \to \infty} \frac{n^2}{1+2+3+\cdots + n} = \lim \limits_{n \to \infty} \frac{n^2}{\frac{n(n+1)}{2}}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ n \text{ in both the numerator and denominator is 2. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  n^2.

\displaystyle  \text{When } n \rightarrow \infty, \text{ then } \frac{1}{n} \rightarrow 0

\displaystyle = \lim \limits_{n \to \infty} \frac{1}{\frac{1(1+\frac{1}{n})}{2}}

\displaystyle = \frac{2}{1} = 2

\\

\displaystyle \text{Question 9: } \lim \limits_{x \to \infty} \frac{3x^{-1}+4x^{-2}}{5x^{-1}+6x^{-2}}

Answer:

\displaystyle \lim \limits_{x \to \infty} \frac{3x^{-1}+4x^{-2}}{5x^{-1}+6x^{-2}} = \frac{\frac{3}{x}+ \frac{4}{x^2}}{\frac{5}{x}+\frac{6}{x^2}} = \frac{3x+4}{5x+6}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0

\displaystyle = \lim \limits_{x \to \infty} \frac{3+\frac{4}{x}}{5+\frac{6}{x}}

\displaystyle = \frac{3}{5}

\\

\displaystyle \text{Question 10: } \lim \limits_{x \to \infty} \frac{  \sqrt{x^2+a^2}  - \sqrt{x^2+b^2} }{\sqrt{x^2+c^2}  - \sqrt{x^2+d^2} }

Answer:

\displaystyle \lim \limits_{x \to \infty} \frac{  \sqrt{x^2+a^2}  - \sqrt{x^2+b^2} }{\sqrt{x^2+c^2}  - \sqrt{x^2+d^2} }

Rationalizing the numerator and denominator we get:

\displaystyle \lim \limits_{x \to \infty} \frac{ ( \sqrt{x^2+a^2}  - \sqrt{x^2+b^2}) }{(\sqrt{x^2+c^2}  - \sqrt{x^2+d^2} )} \times \frac{(\sqrt{x^2+c^2}  + \sqrt{x^2+d^2} )}{(\sqrt{x^2+c^2}  + \sqrt{x^2+d^2} )} \times \frac{( \sqrt{x^2+a^2}  + \sqrt{x^2+b^2})}{( \sqrt{x^2+a^2}  + \sqrt{x^2+b^2})}

\displaystyle \lim \limits_{x \to \infty} \frac{(x^2+a^2) - (x^2+b^2)}{(x^2+c^2) - (x^2+d^2)} \times \frac{(\sqrt{x^2+c^2}  + \sqrt{x^2+d^2} )}{( \sqrt{x^2+a^2}  + \sqrt{x^2+b^2})}

\displaystyle \lim \limits_{x \to \infty} \frac{a^2-b^2}{c^2-d^2} \times \frac{(\sqrt{x^2+c^2}  + \sqrt{x^2+d^2} )}{( \sqrt{x^2+a^2}  + \sqrt{x^2+b^2})}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0

\displaystyle \lim \limits_{x \to \infty} \frac{a^2-b^2}{c^2-d^2} \times \frac{(\sqrt{1+\frac{c^2}{x^2}}  + \sqrt{1+\frac{d^2}{x^2}} )}{( \sqrt{1+\frac{a^2}{x^2}}  + \sqrt{1+\frac{b^2}{x^2}})}

\displaystyle = \frac{a^2-b^2}{c^2-d^2} \times \frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}

\displaystyle = \frac{a^2-b^2}{c^2-d^2}

\\

\displaystyle \text{Question 11: } \lim \limits_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}

Answer:

\displaystyle \lim \limits_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}

\displaystyle = \lim \limits_{n \to \infty} \frac{(n+2)(n+1)!+(n+1)!}{(n+2)(n+1)!-(n+1)!}

\displaystyle = \lim \limits_{n \to \infty} \frac{n+3}{n+1} \times \frac{(n+1)!}{(n+1)!}

\displaystyle = \lim \limits_{n \to \infty} \frac{n+3}{n+1}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ n \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  n.

\displaystyle  \text{When } n \rightarrow \infty, \text{ then } \frac{1}{n} \rightarrow 0

\displaystyle = \lim \limits_{n \to \infty} \frac{1+\frac{3}{n}}{1+\frac{1}{n}}

\displaystyle = \frac{1}{1}

\displaystyle = 1

\\

\displaystyle \text{Question 12: } \lim \limits_{x \to \infty} x \Big\{  \sqrt{x^1+1}- \sqrt{x^2-1} \Big\}

Answer:

\displaystyle \lim \limits_{x \to \infty} x \Big\{  \sqrt{x^1+1}- \sqrt{x^2-1} \Big\}

Rationalizing the numerator:

\displaystyle = \lim \limits_{x \to \infty} x \Big\{  \sqrt{x^1+1}- \sqrt{x^2-1} \Big\} \times \frac{\sqrt{x^1+1}+ \sqrt{x^2-1}}{\sqrt{x^1+1}+ \sqrt{x^2-1}}

\displaystyle = \lim \limits_{x \to \infty} x \Bigg\{  \frac{(x^2+1)-(x^2-1)}{\sqrt{x^1+1}+ \sqrt{x^2-1}} \Bigg\}

\displaystyle = \lim \limits_{x \to \infty}  \Bigg\{  \frac{2x}{\sqrt{x^1+1}+ \sqrt{x^2-1}} \Bigg\}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  x.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x}, \frac{1}{x^2} \rightarrow 0

\displaystyle = \lim \limits_{x \to \infty}  \Bigg\{  \frac{2}{\sqrt{1+\frac{1}{x^2}}+ \sqrt{1-\frac{1}{x^2}}} \Bigg\}

\displaystyle = \frac{2}{\sqrt{1}+\sqrt{1}}

\displaystyle = 1

\\

\displaystyle \text{Question 13: } \lim \limits_{x \to \infty} \Big\{  \sqrt{x+1} - \sqrt{x} \Big\} \sqrt{x+2}

Answer:

\displaystyle \lim \limits_{x \to \infty} \Big\{  \sqrt{x+1} - \sqrt{x} \Big\} \sqrt{x+2}

Rationalizing the numerator:

\displaystyle = \lim \limits_{x \to \infty} \Big\{  \sqrt{x+1} - \sqrt{x} \Big\} \sqrt{x+2} \times \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} + \sqrt{x}}

\displaystyle = \lim \limits_{x \to \infty} \Bigg\{  \frac{(\sqrt{x+2}(x+1-x)}{\sqrt{x+1} + \sqrt{x}}   \Bigg\} 

\displaystyle = \lim \limits_{x \to \infty} \Bigg\{  \frac{\sqrt{x+2}}{\sqrt{x+1} + \sqrt{x}}   \Bigg\} 

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1/2. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  \sqrt{x}.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0

\displaystyle = \lim \limits_{x \to \infty} \Bigg\{  \frac{\sqrt{1+\frac{2}{x}}}{\sqrt{1+\frac{1}{x}} + \sqrt{1}}   \Bigg\} 

\displaystyle = \frac{1}{\sqrt{1}+1} = \frac{1}{2}

\\

\displaystyle \text{Question 14: } \lim \limits_{n \to \infty} \frac{1^2+2^2+ \cdots+ n^2}{n^3}

Answer:

\displaystyle \lim \limits_{n \to \infty} \frac{1^2+2^2+ \cdots+ n^2}{n^3}

\displaystyle = \lim \limits_{n \to \infty} \frac{n(n+1)(2n+1)}{6n^3}

\displaystyle = \lim \limits_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2}

\displaystyle = \lim \limits_{n \to \infty} \frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}

\displaystyle = \frac{2}{6} = \frac{1}{3}

\\

\displaystyle \text{Question 15: } \lim \limits_{n \to \infty} \Bigg(  \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \cdots + \frac{n-1}{n^2}  \Bigg)  

Answer:

\displaystyle \lim \limits_{n \to \infty} \Bigg(  \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \cdots + \frac{n-1}{n^2}  \Bigg)  

\displaystyle = \lim \limits_{n \to \infty} \Bigg(  \frac{1+2+3+ \cdots + (n-1)}{n^2} \Bigg)  

\displaystyle = \lim \limits_{n \to \infty} \Bigg(  \frac{n(n-1)}{2n^2} \Bigg)  

\displaystyle = \lim \limits_{n \to \infty} \Bigg(  \frac{1(1-\frac{1}{n})}{2} \Bigg)  

\displaystyle = \frac{1}{2}

\\

\displaystyle \text{Question 16: } \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{n^4}

Answer:

\displaystyle  \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{n^4}

\displaystyle  = \lim \limits_{n \to \infty} \frac{ \Big(  \frac{n(n+1)}{2} \Big)^2  }{n^4}

\displaystyle  = \lim \limits_{n \to \infty} \frac{  \frac{n^2(n+1)^2}{4}   }{n^4}

\displaystyle  = \lim \limits_{n \to \infty} \frac{n^2}{n^2} \times \frac{(n+1)^2}{4n^2}

\displaystyle  = \lim \limits_{n \to \infty}  \Big(1 + \frac{1}{n} \Big)^2 \times \frac{1}{4}

\displaystyle  = \frac{1}{4}

\\

\displaystyle \text{Question 17: } \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{(n-1)^4}

Answer:

\displaystyle \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{(n-1)^4}

\displaystyle = \lim \limits_{n \to \infty} \frac{  \Big( \frac{n(n+1)}{2} \Big)^2 }{(n-1)^4}

\displaystyle = \lim \limits_{n \to \infty} \frac{ n^2(n+1)^2 }{ 4(n-1)^4 } 

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ n \text{ in both the numerator and denominator is 4. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  n^4.

\displaystyle  \text{When } n \rightarrow \infty, \text{ then } \frac{1}{n} \rightarrow 0

\displaystyle = \lim \limits_{n \to \infty} \frac{ 1(1+\frac{1}{n})^2 }{ 4(1-\frac{1}{n})^4 } 

\displaystyle = \frac{1}{4}

\\

\displaystyle \text{Question 18: } \lim \limits_{x \to \infty} \sqrt{x} \Big\{ \sqrt{x+1}-\sqrt{x}  \Big\}

Answer:

\displaystyle  \lim \limits_{x \to \infty} \sqrt{x} \Big\{ \sqrt{x+1}-\sqrt{x}  \Big\}

\displaystyle  = \lim \limits_{x \to \infty} \sqrt{x} \Bigg\{ (\sqrt{x+1}-\sqrt{x} ) \times \frac{ \sqrt{x+1}+\sqrt{x}}{ \sqrt{x+1}+\sqrt{x}}  \Bigg\}

\displaystyle  = \lim \limits_{x \to \infty} \Bigg\{ \frac{\sqrt{x} (x+1-x)}{\sqrt{x+1}+\sqrt{x}} \Bigg \}

\displaystyle  = \lim \limits_{x \to \infty} \Bigg\{ \frac{\sqrt{x} }{\sqrt{x+1}+\sqrt{x}} \Bigg \}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1/2. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by }  \sqrt{x}.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0

\displaystyle  = \lim \limits_{x \to \infty}  \Bigg\{ \frac{1 }{\sqrt{1+\frac{1}{x}}+1} \Bigg \}

\displaystyle  = \frac{1}{2}

\\

\displaystyle \text{Question 19: } \lim \limits_{n \to \infty} \Bigg(  \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots + \frac{1}{3^n}  \Bigg)  

Answer:

\displaystyle  \lim \limits_{n \to \infty} \Bigg(  \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots + \frac{1}{3^n}  \Bigg)  

\displaystyle  = \lim \limits_{n \to \infty} \Bigg[  \frac{1}{3} \Bigg( 1 + \frac{1}{3^1} + \frac{1}{3^2} + \cdots + \frac{1}{3^(n-1) }  \Bigg) \Bigg] 

\displaystyle  = \lim \limits_{n \to \infty} \Bigg[  \frac{\frac{1}{3} \Big( 1-\frac{1}{3^n} \Big)}{1 - \frac{1}{3}} \Bigg] 

\displaystyle  = \lim \limits_{n \to \infty} \frac{1}{2} \Bigg( 1 - \frac{1}{3^n} \Bigg)

\displaystyle = \frac{1}{2} (1-0) = \frac{1}{2}

\\

\displaystyle \text{Question 20: } \lim \limits_{x \to \infty} \frac{x^4+7x^3+46x+a}{x^4+6} , \text{ where a is a non-zero real number}

Answer:

\displaystyle  \lim \limits_{x \to \infty} \frac{x^4+7x^3+46x+a}{x^4+6}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 4. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by } x^4.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} , \frac{1}{x^2}, \frac{1}{x^3} , \frac{1}{x^4} \rightarrow 0

\displaystyle  = \lim \limits_{x \to \infty} \frac{1+\frac{7}{x}+\frac{46}{x^3}+\frac{a}{x^4} }{1+\frac{6}{x^4} }

\displaystyle  = \frac{1}{1} = 1

\\

\displaystyle \text{Question 21: } f(x) = \frac{ax^2+b}{x^2+1}, \lim \limits_{x \to 0} f(x) = 1 \text{ and } \lim \limits_{x \to \infty} f(x) = 1, \\ \\ \text{ then prove that } f(-2) = f(2) = 1.

Answer:

\displaystyle f(x) = \frac{ax^2+b}{x^2+1}

\displaystyle \lim \limits_{x \to 0} f(x) = 1

\displaystyle \Rightarrow \lim \limits_{x \to 0} \frac{ax^2+b}{x^2+1} = 1

\displaystyle \Rightarrow \frac{a \times 0 + b}{0+1} =1

\displaystyle \Rightarrow b = 1

\displaystyle \text{Also  } \lim \limits_{x \to \infty} f(x) = 1

\displaystyle \Rightarrow \lim \limits_{x \to \infty}\frac{ax^2+b}{x^2+1} = 1

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 2. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by } x^2.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} , \frac{1}{x^2} \rightarrow 0

\displaystyle \Rightarrow \lim \limits_{x \to \infty}\frac{a+\frac{b}{x^2}}{1+\frac{1}{x^2}} = 1

\displaystyle \Rightarrow \frac{a+0}{1+0} = 1

\displaystyle \Rightarrow a = 1

\displaystyle \Rightarrow f(x) = \frac{x^2+1}{x^2+1} = 1

\displaystyle f(x) \text{ is a constant function and does not depend on } x.

\displaystyle \text{Therefore } f(-2) = 1 \text{ and }  f(2) = 1

\\

\displaystyle \text{Question 22: }  \text{Show that } \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - 1) \neq \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x)

Answer:

LHS:

\displaystyle  \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - x)

\displaystyle  = \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - x) \times \frac{\sqrt{x^2+x+1} + x}{\sqrt{x^2+x+1} + x}

\displaystyle  = \lim \limits_{x \to \infty} \frac{x^2+x+1-x^2}{\sqrt{x^2+x+1} + x}

\displaystyle  = \lim \limits_{x \to \infty} \frac{x+1}{\sqrt{x^2+x+1} + x}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by } x.

\displaystyle  \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} , \frac{1}{x^2} \rightarrow 0

\displaystyle  = \lim \limits_{x \to \infty} \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}} + 1}

\displaystyle  = \frac{1}{\sqrt{1}+1} = \frac{1}{2}

RHS:

\displaystyle \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x)

\displaystyle = \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x) \times \frac{\sqrt{x^2+1} +x}{\sqrt{x^2+1} +x}

\displaystyle = \lim \limits_{x \to \infty} \frac{x^2+1-x^2}{\sqrt{x^2+1} +x}

\displaystyle = \lim \limits_{x \to \infty} \frac{1}{\sqrt{x^2+1} +x}

\displaystyle = \frac{1}{\infty} = 0

\displaystyle \text{Therefore }  \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - 1) \neq \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x)

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\displaystyle \text{Question 23: } \lim \limits_{x \to - \infty} ( \sqrt{ 4x^2-7x} +2x)

Answer:

\displaystyle \text{Given: } \lim \limits_{x \to - \infty} ( \sqrt{ 4x^2-7x} +2x)

\displaystyle \text{Let } x = - m. \text{ When } x \rightarrow - \infty, \text{ then } m \rightarrow \infty .

\displaystyle \Rightarrow  \lim \limits_{m \to  \infty} ( \sqrt{ 4m^2+7m} -2m)

\displaystyle =  \lim \limits_{m \to  \infty} ( \sqrt{ 4m^2+7m} -2m) \times \frac{\sqrt{ 4m^2+7m} + 2m}{\sqrt{ 4m^2+7m} +2m}

\displaystyle =  \lim \limits_{m \to  \infty} \frac{4m^2+7m-4m^2}{\sqrt{ 4m^2+7m} + 2m}

\displaystyle =  \lim \limits_{m \to  \infty} \frac{7m}{\sqrt{ 4m^2+7m} + 2m}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ m \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by } m.

\displaystyle  \text{When } m \rightarrow \infty, \text{ then } \frac{1}{m} \rightarrow 0

\displaystyle =  \lim \limits_{m \to  \infty} \frac{7}{\sqrt{ 4+\frac{7}{m}} + 2}

\displaystyle  = \frac{7}{\sqrt{4}+2} = \frac{7}{4}

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\displaystyle \text{Question 24: } \lim \limits_{x \to - \infty} ( \sqrt{ x^2-8x} +x)

Answer:

\displaystyle \text{Given: } \lim \limits_{x \to - \infty} ( \sqrt{ x^2-8x} +x)

\displaystyle \text{Let } x = - m. \text{ When } x \rightarrow - \infty, \text{ then } m \rightarrow \infty .

\displaystyle = \lim \limits_{m \to  \infty} ( \sqrt{ m^2+8m} -m )

\displaystyle = \lim \limits_{m \to  \infty} ( \sqrt{ m^2+8m} -m ) \times \frac{\sqrt{ m^2+8m} +m }{\sqrt{ m^2+8m} + m }

\displaystyle = \lim \limits_{m \to  \infty} \frac{m^2+8m - m^2}{\sqrt{ m^2+8m} + m}

\displaystyle = \lim \limits_{m \to  \infty} \frac{8m}{\sqrt{ m^2+8m} + m}

\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of}  \\  \\ m \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{  in both the numerator and denominator by } m.

\displaystyle  \text{When } m \rightarrow \infty, \text{ then } \frac{1}{m} \rightarrow 0

\displaystyle = \lim \limits_{m \to  \infty} \frac{8}{\sqrt{ 1+\frac{8}{m}} + 1}

\displaystyle = \frac{8}{\sqrt{1}+1} = 4

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\displaystyle \text{Question 25: } \text{Evaluate: } \lim \limits_{n \to \infty} \frac{1^4+2^4+3^4+ \cdots + n^4}{n^5} - \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots + n^3}{n^5}  

Answer:

\displaystyle \text{Consider the identity } (k+1)^5-k^5 = 5k^4+10k^3+10k^2+5k+1

\displaystyle \text{Putting } k = 1, 2, 3 , \cdots , n \text{ and then adding the equations we have }   

\displaystyle (n+1)^4 - 1 = 5 \sum_{k=1}^{n} k^4 + 10 \sum_{k=1}^{n} k^3 + 10 \sum_{k=1}^{n} k^2 + 5 \sum_{k=1}^{n} k + \sum_{k=1}^{n}

\Rightarrow n^5 + 5n^4 + 10n^3 + 10n^2 + 5n = 5 \sum_{k=1}^{n}+ \frac{10n^2(n+1)^2}{4} + \frac{10n(n+1)(2n+1)}{6}+ \frac{5n(n+1)}{2} + n

\Rightarrow 5 \sum_{k=1}^{n} k^4 = n^5+5n^4+10n^3+10n^3+4n - \frac{5n^2(n+1)^2}{2} - \frac{5n(n+1)(2n+1)}{3}- \frac{5n(n+1)}{2}

\displaystyle \Rightarrow 5 \sum_{k=1}^{n} k^4 = n^5 + \frac{5n^4}{2} + \frac{5n^3}{3}- \frac{n}{6}

This expression on further simplification gives

\displaystyle \sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}

\displaystyle \therefore \lim \limits_{n \to \infty} \frac{1^4+2^4+3^4+ \cdots + n^4}{n^5} - \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots + n^3}{n^5}

\displaystyle = \lim \limits_{n \to \infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5} - \lim \limits_{n \to \infty}  \frac{n^2(n+1)^2}{4n^5}

\displaystyle = \frac{1}{30} \lim \limits_{n \to \infty} \Bigg( 1 + \frac{1}{n} \Bigg) \Bigg( 2 + \frac{1}{n} \Bigg) \Bigg(3+\frac{3}{n}- \frac{1}{n^2} \Bigg) - \frac{1}{4} \lim \limits_{n \to \infty} \frac{1}{n} \Bigg(1 + \frac{1}{n} \Bigg)^2

\displaystyle = \frac{1}{30} (1+0)(2+0)(3+0-0)- \frac{1}{4} (0)(1+0)^2

\displaystyle = \frac{1}{5} 

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\displaystyle \text{Question 26: } \text{Evaluate: } \lim \limits_{n \to \infty} \frac{1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1)}{n^3}

Answer:

\displaystyle  \lim \limits_{n \to \infty} \frac{1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1)}{n^3}

\displaystyle  = \lim \limits_{n \to \infty} \frac{ \sum_{k=1}^{n} k ( k+1)    }{n^3}

\displaystyle  = \lim \limits_{n \to \infty} \frac{ \sum_{k=1}^{n} k^2  + \sum_{k=1}^{n} k     }{n^3}

\displaystyle  = \lim \limits_{n \to \infty} \frac{ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}  }{n^3}

\displaystyle  = \lim \limits_{n \to \infty} \frac{n(n+1)(n+2)}{6n^3}

\displaystyle  = \lim \limits_{n \to \infty} \Bigg( 1 + \frac{1}{n} \Bigg) \Bigg( 1 + \frac{2}{n} \Bigg)

\displaystyle  = \frac{1}{3} (1+0) (1+0) = \frac{1}{3}