Tổng hợp câu hay và khó chương 4 - Phần 1

Ta tính giới hạn của các dãy số trong từng đáp án:

+) Đáp án A:           

\(\mathop {\lim }\limits_{} {u_n} = \mathop {\lim }\limits_{} \dfrac{{{{\left( {2017 - n} \right)}^{2018}}}}{{n{{\left( {2018 - n} \right)}^{2017}}}} = \mathop {\lim }\limits_{} \left[ {\dfrac{{2017 - n}}{n}.{{\left( {\dfrac{{2017 - n}}{{2018 - n}}} \right)}^{2017}}} \right]\)

\( = \mathop {\lim }\limits_{} \left[ {\left( {\dfrac{{2017}}{n} - 1} \right){{\left( {\dfrac{{\dfrac{{2017}}{n} - 1}}{{\dfrac{{2018}}{n} - 1}}} \right)}^{2017}}} \right] =  - 1\).

+) Đáp án B:

\(\mathop {\lim }\limits_{} {u_n} = \mathop {\lim }\limits_{} n\left( {\sqrt {{n^2} + 2018}  - \sqrt {{n^2} + 2016} } \right) = \mathop {\lim }\limits_{} \dfrac{{n\left( {{n^2} + 2018 - {n^2} - 2016} \right)}}{{\sqrt {{n^2} + 2018}  + \sqrt {{n^2} + 2016} }}\)

\( = \mathop {\lim }\limits_{} \dfrac{{2n}}{{\sqrt {{n^2} + 2018}  + \sqrt {{n^2} + 2016} }} = \mathop {\lim }\limits_{} \dfrac{2}{{\sqrt {1 + \dfrac{{2018}}{{{n^2}}}}  + \sqrt {1 + \dfrac{{2016}}{{{n^2}}}} }} = 1\).

+) Đáp án C:           

Ta có \({u_{n + 1}} - 1 = \dfrac{1}{2}\left( {{u_n} - 1} \right)\)

\(\begin{array}{l}{u_n} - 1 = \dfrac{1}{2}\left( {{u_{n - 1}} - 1} \right)\\{u_{n - 1}} - 1 = \dfrac{1}{2}\left( {{u_{n - 2}} - 1} \right)\\{u_{n - 2}} - 1 = \dfrac{1}{2}\left( {{u_{n - 3}} - 1} \right)\\...\\{u_2} - 1 = \dfrac{1}{2}\left( {{u_1} - 1} \right)\end{array}\)

\( \Rightarrow \left( {{u_n} - 1} \right)\left( {{u_{n - 1}} - 1} \right)\left( {{u_{n - 2}} - 1} \right)...\left( {{u_2} - 1} \right)\) \( = \dfrac{1}{2}\left( {{u_{n - 1}} - 1} \right).\dfrac{1}{2}\left( {{u_{n - 2}} - 1} \right).\dfrac{1}{2}\left( {{u_{n - 3}} - 1} \right)...\dfrac{1}{2}\left( {{u_1} - 1} \right)\)

\( = \dfrac{1}{{{2^{n - 1}}}}\left( {{u_{n - 1}} - 1} \right)\left( {{u_{n - 2}} - 1} \right)...\left( {{u_2} - 1} \right)\left( {{u_1} - 1} \right)\)

\( \Rightarrow {u_n} - 1 = \dfrac{1}{{{2^{n - 1}}}}\left( {{u_1} - 1} \right)\)

\( \Rightarrow {u_n} = \dfrac{{2016}}{{{2^{n - 1}}}} + 1 \Leftrightarrow {u_n} = 4032.{\left( {\dfrac{1}{2}} \right)^n} + 1\)\( \Rightarrow \mathop {\lim }\limits_{} {u_n} = 1\).

Ở câu C, các em cũng có thể giả sử \(\lim {u_n} = a\) thì \({u_{n + 1}} \to a\) khi \(n \to \infty \), do đó ta có \(a = \dfrac{1}{2}\left( {a + 1} \right) \Leftrightarrow a = 1\)

+) Đáp án D:

Ta có \({u_n} = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}} = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{n} - \dfrac{1}{{n + 1}} = 1 - \dfrac{1}{{n + 1}} = \dfrac{n}{{n + 1}}\)

\( \Rightarrow \mathop {\lim }\limits_{} {u_n} = \mathop {\lim }\limits_{} \dfrac{n}{{n + 1}} = 1\).

Ta có $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{2016}} + x - 2}}{{\sqrt {2018x + 1}  - \sqrt {x + 2018} }}$ $ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {{x^{2016}} + x - 2} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{2018x + 1 - x - 2018}}$

\( = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {{x^{2015}} + {x^{2014}} + ... + x + 2} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{2017\left( {x - 1} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {{x^{2015}} + {x^{2014}} + ... + x + 2} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{2017}} = 2\sqrt {2019} \)

Mà \(f\left( 1 \right) = k\)

Suy ra hàm số liên tục tại \(x = 1\) \( \Leftrightarrow k = 2\sqrt {2019} \).

Vì hàm số có giới hạn hữu hạn tại $x = 1$ nên biểu thức tử nhận $x = 1$ làm nghiệm, hay $1 + a + b = 0$.

Áp dụng vào giả thiết, được $\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + ax - 1 - a}}{{{x^2} - 1}} = \dfrac{{ - 1}}{2} \Leftrightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 1 + a} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} =  - \dfrac{1}{2}$.

$ \Leftrightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 1 + a}}{{x + 1}} =  - \dfrac{1}{2} \Leftrightarrow \dfrac{{2 + a}}{2} =  - \dfrac{1}{2} \Leftrightarrow a =  - 3$.

Suy ra $b = 2$.

Vậy ${a^2} + {b^2} = 13$.

Bước 1:

Ta có: \(f\left( { - 2} \right) =  - 11\), $\mathop {\lim }\limits_{x \to  - {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {2^ - }} \left( {3x - 5} \right) =  - 11$, $\mathop {\lim }\limits_{x \to  - {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {2^ - }} \left( {ax - 1} \right) =  - 2a - 1$.

Bước 2:

Để hàm số liên tục tại \(x =  - 2\) thì $f\left( { - 2} \right) = \mathop {\lim }\limits_{x \to  - {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {2^ + }} f\left( x \right)$

$ \Leftrightarrow  - 2a - 1 =  - 11 \Leftrightarrow a = 5$.

Vậy hàm số liên tục tại \(x =  - 2\) khi \(a = 5\).

Cách 1 (Đặc biệt hóa)

Chọn \(f\left( x \right) = 10x\), ta có \(\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 20}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{10x - 20}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{10\left( {x - 2} \right)}}{{x - 2}} = 10\)

Lúc đó \(T = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{60x + 5}} - 5}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{60x + 5}} - 5}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 2} \dfrac{{60x + 5 - {5^3}}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {{{\sqrt[3]{{60x + 5}}}^2} + 5\sqrt[3]{{60x + 5}} + 25} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 2} \dfrac{{60\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {{{\sqrt[3]{{60x + 5}}}^2} + 5\sqrt[3]{{60x + 5}} + 25} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 2} \dfrac{{60}}{{\left( {x + 3} \right)\left( {{{\sqrt[3]{{60x + 5}}}^2} + 5\sqrt[3]{{60x + 5}} + 25} \right)}} = \dfrac{4}{{25}}\)

Cách 2:

Chọn \(f\left( x \right) = 10x\), ta có \(\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 20}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{10x - 20}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{10\left( {x - 2} \right)}}{{x - 2}} = 10\)

Sử dụng CASIO (chức năng CALC), nhập hàm cần tính giới hạn

Màn hình hiển thị

Thay giá trị \(x = 1,9999999\) vào

Màn hình hiển thị

Thay tiếp giá trị \(x = 2,0000001\) vào

Màn hình hiển thị

Cách 3:

Theo giả thiết có \(\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) - 20} \right) = 0\) hay \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 20\) \(\left( * \right)\)

Khi đó $T = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{6f\left( x \right) + 5 - 125}}{{\left( {{x^2} + x - 6} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right) + 25} \right]}}$

$T = \mathop {\lim }\limits_{x \to 2} \dfrac{{6\left[ {f\left( x \right) - 20} \right]}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right) + 25} \right]}}$

\(\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 20}}{{x - 2}} = 10\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 20\\ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{6}{{\left( {x - 3} \right).\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right) + 25} \right]}}\\ = \dfrac{6}{{\left( {2 + 3} \right).\left[ {{{\left( {\sqrt[3]{{6.20 + 5}}} \right)}^2} + 5.\left( {\sqrt[3]{{6.20 + 5}}} \right) + 25} \right]}} = \dfrac{6}{{5.75}}\end{array}\)

$T = \dfrac{{10.6}}{{5.75}} = \dfrac{4}{{25}}$

Đặt \(f\left( x \right) = {x^3} - 3{x^2} + \left( {2m - 2} \right)x + m - 3\). Ta thấy hàm số liên tục trên \(\mathbb{R}\).

Dễ thấy nếu \(x \to  - \infty \)thì \(f\left( x \right) \to  - \infty \) hay \(f\left( x \right) < 0\)

Suy ra điều kiện cần để \(f\left( x \right) = 0\) có \(3\) nghiệm thỏa \({x_1} <  - 1 < {x_2} < {x_3}\) là $f\left( { - 1} \right) > 0 \Leftrightarrow  - m - 5 > 0 \Leftrightarrow m <  - 5$.

Điều kiện đủ: với \(m <  - 5\) ta có

*) \(\mathop {\lim }\limits_{x \to  - \infty } f\left( x \right) =  - \infty \) nên tồn tại \(a <  - 1\) sao cho \(f\left( a \right) < 0\)

Mặt khác \(f\left( { - 1} \right) =  - m - 5 > 0\). Suy ra \(f\left( a \right).f\left( { - 1} \right) < 0\).

Do đó tồn tại \({x_1} \in \left( {a; - 1} \right)\) sao cho \(f\left( {{x_1}} \right) = 0\).

*) \(f\left( 0 \right) = m - 3 < 0\), \(f\left( { - 1} \right) > 0\). Suy ra \(f\left( 0 \right).f\left( { - 1} \right) < 0\).

Do đó tồn tại \({x_2} \in \left( { - 1;0} \right)\) sao cho \(f\left( {{x_2}} \right) = 0\).

*) \(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) =  + \infty \) nên tồn tại \(b > 0\) sao cho \(f\left( b \right) > 0\)

Mặt khác \(f\left( 0 \right) < 0\). Suy ra \(f\left( 0 \right).f\left( b \right) < 0\).

Do đó tồn tại \({x_3} \in \left( {0;b} \right)\) sao cho \(f\left( {{x_3}} \right) = 0\).

Vậy \(m <  - 5\) thỏa mãn yêu cầu bài toán.

Đặt \(L = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{x}{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - 2}}} \right) = \dfrac{a}{b}\) thì \(\dfrac{1}{L} = \lim \left( {\dfrac{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - 2}}{x}} \right) = \dfrac{b}{a}\).

Ta có

$\dfrac{b}{a} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - \sqrt {x + 4}  + \sqrt {x + 4}  - 2}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt[7]{{x + 1}}.\sqrt {x + 4}  - \sqrt {x + 4} }}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {x + 4}  - 2}}{x}} \right)$

Xét ${L_1} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {x + 4} \left( {\sqrt[7]{{x + 1}} - 1} \right)}}{x}} \right)$. Đặt \(t = \sqrt[7]{{x + 1}}\). Khi đó :\(\left\{ \begin{array}{l}x = {t^7} - 1\\x \to 0 \Rightarrow t \to 1\end{array} \right.\)

\({L_1} = \mathop {\lim }\limits_{t \to 1} \dfrac{{\sqrt {{t^7} + 3} \left( {t - 1} \right)}}{{{t^7} - 1}} = \mathop {\lim }\limits_{t \to 1} \dfrac{{\sqrt {{t^7} + 3} }}{{\left( {{t^6} + {t^5} + {t^4} + {t^3} + {t^2} + t + 1} \right)}} = \dfrac{2}{7}\)

Xét ${L_2} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {x + 4}  - 2}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {x + 4}  - 2} \right)\left( {\sqrt {x + 4}  + 2} \right)}}{{x\left( {\sqrt {x + 4}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\sqrt {x + 4}  + 2}} = \dfrac{1}{4}$

Vậy \(\dfrac{b}{a} = \dfrac{2}{7} + \dfrac{1}{4} = \dfrac{{15}}{{28}}\)\( \Rightarrow a = 28,b = 15 \Rightarrow a + b = 43\) \( \Rightarrow a + b = 43\).

Ta có

\(\begin{array}{l}{u_2} = {u_1} + 4.1 + 3\\{u_3} = {u_2} + 4.2 + 3\\...\\{u_n} = {u_{n - 1}} + 4.\left( {n - 1} \right) + 3\end{array}\)

Cộng vế theo vế và rút gọn ta được

\({u_n} = {u_1} + 4.\left( {1 + 2 + ... + n - 1} \right) + 3\left( {n - 1} \right)\)\( = 4\dfrac{{n\left( {n - 1} \right)}}{2} + 3\left( {n - 1} \right)\)\( = 2{n^2} + n - 3\), với mọi $n \ge 1$.

Suy ra                                                           

\(\begin{array}{l}{u_{2n}} = 2{\left( {2n} \right)^2} + 2n - 3\\{u_{{2^2}n}} = 2{\left( {{2^2}n} \right)^2} + {2^2}n - 3\\...\\{u_{{2^{2018}}n}} = 2{\left( {{2^{2018}}n} \right)^2} + {2^{2018}}n - 3\end{array}\)

Và

\(\begin{array}{l}{u_{4n}} = 2{\left( {4n} \right)^2} + 4n - 3\\{u_{{4^2}n}} = 2{\left( {{4^2}n} \right)^2} + {4^2}n - 3\\...\\{u_{{4^{2018}}n}} = 2{\left( {{4^{2018}}n} \right)^2} + {4^{2018}}n - 3\end{array}\)

Do đó \(\lim \dfrac{{\sqrt {{u_n}}  + \sqrt {{u_{4n}}}  + \sqrt {{u_{{4^2}n}}}  + ... + \sqrt {{u_{{4^{2018}}n}}} }}{{\sqrt {{u_n}}  + \sqrt {{u_{2n}}}  + \sqrt {{u_{{2^2}n}}}  + ... + \sqrt {{u_{{2^{2018}}n}}} }}\)

$ = \lim \dfrac{{\sqrt {2 + \dfrac{1}{n} - \dfrac{3}{{{n^2}}}}  + \sqrt {{{2.4}^2} + \dfrac{4}{n} - \dfrac{3}{{{n^2}}}}  + ... + \sqrt {2{{\left( {{4^{2018}}} \right)}^2} + \dfrac{{{4^{2018}}}}{n} - \dfrac{3}{{{n^2}}}} }}{{\sqrt {2 + \dfrac{1}{n} - \dfrac{3}{{{n^2}}}}  + \sqrt {{{2.2}^2} + \dfrac{2}{n} - \dfrac{3}{{{n^2}}}}  + ... + \sqrt {2{{\left( {{2^{2018}}} \right)}^2} + \dfrac{{{2^{2018}}}}{n} - \dfrac{3}{{{n^2}}}} }}$

\( = \dfrac{{\sqrt 2 \left( {1 + 4 + {4^2} + ... + {4^{2018}}} \right)}}{{\sqrt 2 \left( {1 + 2 + {2^2} + ... + {2^{2018}}} \right)}}\)$ = \dfrac{{1\dfrac{{1 - {4^{2019}}}}{{1 - 4}}}}{{\dfrac{{1 - {2^{2019}}}}{{1 - 2}}}}$$ = \dfrac{1}{3}\dfrac{{{4^{2019}} - 1}}{{{2^{2019}} - 1}}$\( = \dfrac{{{2^{2019}} + 1}}{3}\).

Vì \({2^{2019}} > 2019\) cho nên sự xác định ở trên là duy nhất nên \(\left\{ \begin{array}{l}a = 2\\b = 1\\c = 3\end{array} \right.\)

Vậy \(S = a + b - c = 0\).

Ta có \(\dfrac{1}{{n\sqrt {n + 1}  + \left( {n + 1} \right)\sqrt n }}\)\( = \dfrac{1}{{\sqrt n \sqrt {n + 1} \left( {\sqrt n  + \sqrt {n + 1} } \right)}}\) \( = \dfrac{{\sqrt {n + 1}  - \sqrt n }}{{\sqrt n \sqrt {n + 1} }} = \dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}\)

Suy ra

\({S_n} = \dfrac{1}{{1\sqrt 2  + 2\sqrt 1 }} + \dfrac{1}{{2\sqrt 3  + 3\sqrt 2 }} + ... + \dfrac{1}{{n\sqrt {n + 1}  + \left( {n + 1} \right)\sqrt n }}\)

\( = \dfrac{1}{1} - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 3 }} + ... + \dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }} = 1 - \dfrac{1}{{\sqrt {n + 1} }}\)

Suy ra \(\lim {S_n} = 1\)

Cách 1:

Xét dãy số \(\left( {{u_n}} \right)\), với \({u_n} = \left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)\), \(n \ge 2,\,n \in \mathbb{N}\).

Ta có:

\({u_2} = 1 - \dfrac{1}{{{2^2}}} = \dfrac{3}{4} = \dfrac{{2 + 1}}{{2.2}}\);

\({u_3} = \left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right) = \dfrac{3}{4}.\dfrac{8}{9} = \dfrac{4}{6} = \dfrac{{3 + 1}}{{2.3}}\);

\({u_4} = \left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)\left( {1 - \dfrac{1}{{{4^2}}}} \right) = \dfrac{3}{4}.\dfrac{8}{9}.\dfrac{{15}}{{16}} = \dfrac{5}{8} = \dfrac{{4 + 1}}{{2.4}}\)

\( \cdots  \cdots \)

\({u_n} = \dfrac{{n + 1}}{{2n}}\).

Dễ dàng chứng minh bằng phương pháp qui nạp để khẳng định \({u_n} = \dfrac{{n + 1}}{{2n}},\,\forall n \ge 2\)

Khi đó \(\lim \left[ {\left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)} \right] = \lim \dfrac{{n + 1}}{{2n}} = \dfrac{1}{2}\).

$I = \mathop {\lim }\limits_{x \to 1} \dfrac{{2x - \sqrt {x + 3} }}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - \sqrt {x + 3} } \right)\left( {2x + \sqrt {x + 3} } \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + \sqrt {x + 3} } \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{4{x^2} - x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + \sqrt {x + 3} } \right)}}$

$ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {4x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + \sqrt {x + 3} } \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{4x + 3}}{{\left( {x + 1} \right)\left( {2x + \sqrt {x + 3} } \right)}} = \dfrac{7}{8}$.

Ta có \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - 2 + 2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - 2}}{{\sqrt 2 \left( {x - 1} \right)}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = I + J\).

Tính \(I = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - 2}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + x + 2 - 4}}{{\sqrt 2 \left( {x - 1} \right)\left( {\sqrt {{x^2} + x + 2}  + 2} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\sqrt 2 \left( {x - 1} \right)\left( {\sqrt {{x^2} + x + 2}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 2}}{{\sqrt 2 \left( {\sqrt {{x^2} + x + 2}  + 2} \right)}} = \dfrac{3}{{4\sqrt 2 }}\).

và \(J = \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{8 - 7x - 1}}{{\sqrt 2 \left( {x - 1} \right)\left[ {4 + 2\sqrt[3]{{7x + 1}} + {{\left( {\sqrt[3]{{7x + 1}}} \right)}^2}} \right]}}\)

\(\mathop { = \lim }\limits_{x \to 1} \dfrac{{ - 7}}{{\sqrt 2 \left[ {4 + 2\sqrt[3]{{7x + 1}} + {{\left( {\sqrt[3]{{7x + 1}}} \right)}^2}} \right]}} = \dfrac{{ - 7}}{{12\sqrt 2 }}\).

Do đó \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = I + J = \dfrac{{\sqrt 2 }}{{12}}\)

Suy ra \(a = 1\), \(b = 12\), \(c = 0\). Vậy \(a + b + c = 13\).