11类RD Sharma解决方案–第29章限制–练习29.4 |套装2

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问题18. Lim _x→1 {√(5x – 4)–√x} /(x ³ – 1)

解决方案：

We have, Lim_x→1{√(5x – 4) – √x}/(x³ – 1)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→1}\frac{\sqrt{(5x - 4)} - \sqrt{x}}{(x^3 - 1)} \times \frac{\sqrt{(5x - 4)} + \sqrt{x}}{\sqrt{(5x - 4)} + \sqrt{x}}$

= Lim_x→1{(5x – 4) – x}/[{√(5x – 4) + √x}(x³ – 1)]

= Lim_x→1{4(x – 1)}/[{√(5x – 4) + √x}(x-1)(x² + x + 1)]

= Lim_x→1(4)/[{√(5x – 4) + √x}(x² + x + 1)]

Now put x = 1, we get

= 4/{(3)(√1 + √1)}

= 4/6

= 2/3

问题19. Lim _x→2 {√(1 + 4x)–√(5 + 2x)} /(x – 2)

解决方案：

We have, Lim_x→2{√(1 + 4x) – √(5 + 2x)}/(x – 2)

Find the limit of the given equation

When we put x = 2, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→2}\frac{\sqrt{(1 + 4x)} - \sqrt{(5 + 2x)}}{(x - 2)} \times \frac{\sqrt{(1 + 4x)} + \sqrt{(5 + 2x)}}{\sqrt{(1 + 4x)} + \sqrt{(5 + 2x)}}$

= Lim_x→2{√(1 + 4x) – √(5+2x)}/[(x – 2){√(1 + 4x) + √(5 + 2x)}]

= Lim_x→2{(1 + 4x) – (5 + 2x)}/[(x – 2){√(1 + 4x) + √(5 + 2x)}]

= Lim_x→2{2(x – 2)}/[(x – 2){√(1 + 4x) + √(5 + 2x)}]

= Lim_x→2(2)/{√(1 + 4x) + √(5 + 2x)}

Now put x = 2, we get

= 2/{√(1 + 8) + √(5 + 4)}

= 2/(3 + 3)

= 1/3

问题20. Lim _x→1 {√(3 + x)–√(5 – x)} /(x ² – 1)

解决方案：

We have, Lim_x→1{√(3 + x) – √(5 – x)}/(x² – 1)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→1}\frac{\sqrt{(3 + x)} - \sqrt{(5 - x)}}{(x^2 - 1)} \times \frac{\sqrt{(3 + x)} + \sqrt{(5 - x)}}{\sqrt{(3 + x)} + \sqrt{(5 - x)}}$

= Lim_x→1{(3 + x) – (5 – x)}/[(x² – 1){√(3 + x) + √(5 – x)}]

= Lim_x→1{2(x – 1)}/[(x – 1)(x + 1){√(3 + x) + √(5 – x)}]

= Lim_x→1(2)/[(x + 1){√(3 + x) + √(5 – x)}]

Now put x = 1, we get

= 2/{2(2 + 2)}

= 1/4

问题21. Lim _x→0 {√(1 + x ² )–√(1 – x ² )} /(x)

解决方案：

We have, Lim_x→0{√(1 + x²) – √(1 – x²)}/(x)

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→0}\frac{\sqrt{(1 + x^2)} - \sqrt{(1 - x^2)}}{x} \times \frac{\sqrt{(1 + x^2)} + \sqrt{(1 - x^2)}}{\sqrt{(1 + x^2)} + \sqrt{(1 - x^2)}}$

= Lim_x→0{(1 + x²) – (1 – x²)}/[x{√(1 + x²) + √(1 – x²)}]

= Lim_x→0{(1 + x²) – (1 – x²)}/[x{√(1 + x²) + √(1 – x²)}]

= Lim_x→0(2x²/[x{√(1 + x²) + √(1 – x²)}]

= Lim_x→0(2x/{√(1 + x²) + √(1 – x²)}

Now put x = 0, we get

= 2 × 0/(√1 + √1)

= 0

问题22. Lim _x→0 {√(1 + x + x ² )–√(x + 1)} /(2x ² )

解决方案：

We have, Lim_x→0{√(1 + x + x²) – √(x + 1)}/(2x²)

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→0}\frac{\sqrt{(1 + x + x^2)} - \sqrt{(x + 1)}}{2x^2} \times \frac{\sqrt{(1 + x + x^2)} + \sqrt{(x + 1)}}{\sqrt{(1 + x + x^2)} + \sqrt{(x + 1)}}$

= Lim_x→0{(1 + x + x²) – (x + 1)}/[2x²{√(1 + x + x²) – √(x + 1)}]

= Lim_x→0(x²)/[2x²{√(1 + x + x²) – √(x + 1)}]

= Lim_x→0(1)/[2{√(1 + x + x²) – √(x + 1)}]

Now put x = 0, we get

= 1/{2(√1 + √1)

= 1/4

问题23. Lim _x→4 {2 –√x} /(4 – x)

解决方案：

We have, Lim_x→4{2 – √x}/(4 – x)

Find the limit of the given equation

When we put x = 4, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→4}\frac{2 - √x}{(4 - x)} \times \frac{2 + √x}{2 + √x}$

= Lim_x→4{4 – x}/[(4 – x){2 + √x}]

= Lim_x→4{4 – x}/[(4 – x){2 + √x}]

= Lim_x→4(1)/{2 + √x}

Now put x = 4, we get

= 1/(2 + 2)

= 1/4

问题24. Lim _x→a (x – a)/ {√x–√a}

解决方案：

We have, Lim_x→a(x – a)/{√x – √a}

Find the limit of the given equation

When we put x = a, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

=  $Lim_{x→a}\frac{(x - a)}{√x - √a} \times \frac{√x + √a}{√x + √a}$

= Lim_x→a[(x – a){√x – √a}]/(x – a)

= Lim_x→a{√x + √a}

Now put x = a, we get

= √a + √a

= 2√a

问题25. Lim _x→0 {√(1 + 3x)–√(1 – 3x)} /(x)

解决方案：

We have, Lim_x→0{√(1 + 3x) – √(1 – 3x)}/(x)

Find the limit of the given equation

When we put x = a, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→0}\frac{\sqrt{(1 + 3x)} - \sqrt{(1 - 3x)}}{x} \times \frac{\sqrt{(1 + 3x)} + \sqrt{(1 - 3x)}}{\sqrt{(1 + 3x)} + \sqrt{(1 - 3x)}}$

= Lim_x→0{(1 + 3x) – (1 – 3x)}/[(x){√(1 + 3x) – √(1 – 3x)}]

= Lim_x→0(6x)/[(x){√(1 + 3x) – √(1 – 3x)}]

= Lim_x→0(6)/{√(1 + 3x) – √(1 – 3x)}

Now put x = 0, we get

= 6/(√1 + √1)

= 6/2

= 3

问题26. Lim _x→0 {√(2 – x)–√(2 + x)} /(x)

解决方案：

We have, Lim_x→0{√(2 – x) – √(2 + x)}/(x)

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→0}\frac{\sqrt{(2 - x)} - \sqrt{(2 + x)}}{x} \times \frac{\sqrt{(2 - x)} + \sqrt{(2 + x)}}{\sqrt{(2 - x)} + \sqrt{(2 + x)}}$

= Lim_x→0{(2 – x) – (2 + x)}/[x{√(2 – x) + √(2 + x)}]

= Lim_x→0(-2x)/[x{√(2 – x) + √(2 + x)}]

= Lim_x→0(-2)/{√(2 – x) + √(2 + x)}

Now put x = 0, we get

= (-2)/(√2 + √2)

= (-2)/(2√2)

= -1/(√2)

问题27.林_x→1 {√(3 + x)–√(5 – x)} /(x ² – 1)

解决方案：

We have, Lim_x→1{√(3 + x) – √(5 – x)}/(x² – 1)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{x→1}\frac{\sqrt{(3 + x)} - \sqrt{(5 - x)}}{(x^2 - 1)} \times \frac{\sqrt{(3 + x)} + \sqrt{(5 - x)}}{\sqrt{(3 + x)} + \sqrt{(5 - x)}}$

= Lim_x→1{(3 + x) – (5 – x)}/[(x² – 1){√(3 + x) + √(5 – x)}]

= Lim_x→1{2(x – 1)}/[(x – 1)(x + 1){√(3 + x) + √(5 – x)}]

= Lim_x→1(2)/[(x + 1){√(3 + x) + √(5 – x)}]

Now put x = 1, we get

= 2/{(2)(√4 + √4)}

= 2/8

= 1/4

问题28.林_x→1 {(2x – 3)(√x– 1)} /(3x ² + 3x – 6)

解决方案：

We have, Lim_x→1{(2x – 3)(√x – 1)}/(3x² + 3x – 6)

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Lim_x→1{(2x – 3)(x – 1)}/[(3x² + 3x – 6)(√x + 1)]

= Lim_x→1{(2x – 3)(x – 1)}/[3(x² + x – 2)(√x + 1)]

= Lim_x→1{(2x – 3)(x – 1)}/[3(x – 1)(x + 2)(√x + 1)]

= Lim_x→1(2x – 3)/[3(x + 2)(√x + 1)]

Now put x = 1, we get

= (2 – 3)/{3(3)(√1 + 1)

= -1/(3 × 3 × 2)

= -1/18

问题29. Lim _x→0 {√(1 + x ² )–√(1 + x)} / {√(1 + x ³ )–√(1 + x)}

解决方案：

We have, Lim_x→0{√(1 + x²) – √(1 + x)}/{√(1 + x³) – √(1 + x)}

Find the limit of the given equation

When we put x = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

$=\lim_{x\to0}\frac{(\sqrt{1+x^2}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})}$

$=\lim_{x\to0}\frac{x(x-1)}{x(x^2-1)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}$

$=\lim_{x\to0}\frac{[(1+x^2)-(1+x)]}{[(1+x^3)-(1+x)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}$

$=\lim_{x\to0}\frac{(x^2-x)}{(x^3-x)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}$

$=\lim_{x\to0}\frac{1}{(x+1)}×\frac{(\sqrt{1+x^3}+\sqrt{x+1})}{(\sqrt{1+x^2}+\sqrt{x+1})}$

Now put x = 0, we get

= (√1 + √1)/{1(√1 + √1)}

= 2/2

= 1

问题30.林_x→1 {x ² –√x} / {√x– 1}

解决方案：

We have, Lim_x→1{x² – √x}/{√x – 1}

Find the limit of the given equation

When we put x = 1, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= Lim_x→1{√x(x√x -1)}/{√x – 1}

= Lim_x→1{√x(x^3/2 – 1)}/{√x – 1}

= Lim_x→1[√x{(√x)³ – 1}]/{√x – 1}

= Lim_x→1[(√x)(√x – 1)(x + √x + 1)]/{√x – 1}

= Lim_x→1[(√x)(x + √x + 1)]

Now put x = 1, we get

= (√1)(1 + √1 + 1)

= 3

问题31. Lim _h→0 {√(x + h)–√x} /(h)，x≠0

解决方案：

We have, Lim_h→0{√(x + h) – √x}/(h)

Find the limit of the given equation

When we put h = 0, this expression takes the form of 0/0.

So, on rationalizing the given equation we get

= $Lim_{h→0}\frac{\sqrt{(x + h)} - \sqrt{x}}{h} \times \frac{\sqrt{(x + h)} + \sqrt{x}}{\sqrt{(x + h)} + \sqrt{x}}$

= Lim_h→0{(x + h) – x}/[h{√(x + h) + √x}]

= Lim_h→0(h)/[h{√(x + h) + √x}]

= Lim_h→0(1)/{√(x + h) + √x}

Now put x = 0, we get

= 1/(√x + √x)

= 1/(2√x)

问题32.林_x→√10 {√(7 + 2x)–(√5+√2)} /(x ² – 10)

解决方案：

We have, Lim_x→√10{√(7 + 2x) – (√5 + √2)}/(x² – 10)

= Lim_x→√10{√(7 + 2x) – √(√5 + √2)²}/{(x – √10)(x + √10)}

= Lim_x→√10{√(7 + 2x) – √(5 + 2 + 2√5√2)}/{(x – √10)(x + √10)}

= Lim_x→√10{√(7 + 2x) – √(7 + 2√10)}/{(x – √10)(x + √10)}

On rationalizing numerator.

=$\lim_{x\to\sqrt{10}}\frac{(\sqrt{7+2x}-\sqrt{7+2\sqrt{10}})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

=$\lim_{x\to\sqrt{10}}\frac{(7+2x)-(7+2\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

=$\lim_{x\to\sqrt{10}}\frac{2(x-\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

Now put x = √10, we get

=$\frac{2}{(\sqrt{10}+\sqrt{10})(2\sqrt{7+2\sqrt{10}}}$

=$\frac{1}{(2\sqrt{10})(\sqrt{7+2\sqrt{10}}}$

=$\frac{1}{(2\sqrt{10})(\sqrt{\sqrt{5}+\sqrt{2})^2}}$

= 1/{(2√10)(√5 + √2)}

On rationalizing denominator.

= (√5 – √2)/{(2√10)(5 – 2)}

= (√5 – √2)/(6√10)

问题33.林_x→√6 {√(5 + 2x)–(√3+√2)} /(x ² – 6)

解决方案：

We have, Lim_x→√6{√(5 + 2x) – (√3 + √2)}/(x² – 6)

= Lim_x→√6{√(5 + 2x) – √(√3 + √2)²}/{(x – √6)(x + √6)}

= Lim_x→√6{√(5 + 2x) – √(3 + 2 + 2√3√2)}/{(x – √6)(x + √6)}

= Lim_x→√6{√(5 + 2x) – √(5 + 2√6)}/{(x  -√6)(x + √6)}

On rationalizing numerator.

= $\lim_{x\to\sqrt{6}}\frac{(\sqrt{5+2x}-\sqrt{5+2\sqrt{6}})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

=$\lim_{x\to\sqrt{6}}\frac{(5+2x)-(5+2\sqrt{6})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

=$\lim_{x\to\sqrt{6}}\frac{2(x-\sqrt{6})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

=$\lim_{x\to\sqrt{6}}\frac2{(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

Now put x = √6, we get

=$\frac2{(\sqrt6+\sqrt{6})(\sqrt{5+2\sqrt{5}}+\sqrt{5+2\sqrt{6}})}$

=$\frac2{(2\sqrt{6})(2\sqrt{5+2\sqrt{5}})}$

=$\frac1{(2\sqrt{6})(\sqrt{5+2\sqrt{5}})}$

= 1/{(2√6)(√3 + √2)}

On rationalizing denominator, we get

= (√3 – √2)/{(2√6)(3 – 2)}

= (√3 – √2)/(2√6)

问题34. Lim _x→√2 {√(3 + 2x)–(√2+1)} /(x ² – 2)

解决方案：

We have,  Lim_x→√2{√(3 + 2x) – (√2 + 1)}/(x² – 2)

= Lim_x→√2{√(3 + 2x) – √(√2 + 1)²}/{(x – √2)(x + √2)}

= Lim_x→√2{√(3 + 2x) – √(2 + 1 + 2√3)}/{(x – √2)(x + √2)}

= Lim_x→√2{√(3 + 2x) – √(3 + 2√3)}/{(x – √2)(x + √2)}

On rationalizing numerator.

=$\lim_{x\to\sqrt{2}}\frac{(\sqrt{3+2x}-\sqrt{3+2\sqrt{2}})(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}})}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}})}$

=$\lim_{x\to\sqrt{2}}\frac{(3+2x)-(3+2\sqrt2)}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}$

=$\lim_{x\to\sqrt{2}}\frac{(3+2x)-(3+2\sqrt2)}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}$

=$\lim_{x\to\sqrt{2}}\frac{2(x-\sqrt{3})}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}$

=$\lim_{x\to\sqrt{2}}\frac{2}{(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt2})}$

Now put x = √2, we get

=$\frac2{(\sqrt2+\sqrt2)(\sqrt{3+2\sqrt{5}}+\sqrt{3+2\sqrt2})}$

=$\frac2{(2\sqrt2)(2\sqrt{3+2\sqrt2})}$

=$\frac1{(2\sqrt{2})(\sqrt{3+2\sqrt2})}$

= 1/{(2√2)(√2 + 1)}

On rationalizing denominator, we get

= (√2 – 1)/{(2√2)(2 – 1)}

= (√2 – 1)/(2√2)