Trắc nghiệm Ôn tập Chương IV-Giới hạn (có đáp án)

Ta có: $$\begin{gathered} u_n=\sqrt{n^2+1}-n=\frac{\left(\sqrt{n^2+1}-n\right)\left(\sqrt{n^2+1}+n\right)}{\sqrt{n^2+1}+n} \\ =\frac{1}{n\left(\sqrt{1+\frac{1}{n^2}}+1\right)}=\frac{1}{n}.\frac{1}{\sqrt{1+\frac{1}{n^2}}+1} \end{gathered}$$

Vì $\lim \frac{1}{n}=0,\lim \frac{1}{\sqrt{1+\frac{1}{n^2}}+1}=\frac{1}{2}$ nên $\lim u_n=0$.

Chọn đáp án A.

Ta có: $u_n=\frac{-2n^3+3n^2+4}{n^4+4n^3+n}=\frac{\frac{-2n^3+3n^2+4}{n^4}}{\frac{n^4+4n^3+n}{n^4}}=\frac{-\frac{2}{n}+\frac{3}{n^2}+\frac{4}{n^4}}{1+\frac{4}{n}+\frac{1}{n^3}}$

Mà $\lim \frac{2}{n}=0, \lim \frac{3}{n^2}=0, \lim \frac{4}{n^4}=0, \lim \frac{4}{n}=0 và \lim \frac{1}{n^3}=0$

Do đó $\lim u_n=\frac{0+0+0}{1+0+0}=0$

Chọn đáp án B

Ta có: $u_n=\frac{{\left(-1\right)}^n{.2}^{5n+1}}{3^{5n+2}}=\frac{{\left(-1\right)}^n{\mathrm{.2.2}}^{5n}}{3^2{.3}^{5n}}=\frac{\left(-1\right).2}{9}.{\left(\frac{2}{3}\right)}^{5n}$

Vì $\left|\frac{2}{3}\right|<1$ nên $\lim {\left(\frac{2}{3}\right)}^{5n}=0$. Do đó $\lim u_n=0$

Chọn đáp án B

Ta có: $u_n=\frac{{\left(-5\right)}^n\left(1+\frac{4^n}{{\left(-5\right)}^n}\right)}{{\left(-7\right)}^n\left(-7+\frac{{4.4}^n}{{\left(-7\right)}^n}\right)}={\left(\frac{5}{7}\right)}^n\cdot \left(\frac{1+{\left(-\frac{4}{5}\right)}^n}{-7+4.{\left(\frac{-4}{7}\right)}^n}\right)$

Vì $\lim {\left(-\frac{4}{5}\right)}^n=\lim {\left(-\frac{4}{7}\right)}^n=0$ nên $\lim \left(\frac{1+{\left(-\frac{4}{5}\right)}^n}{-7+4.{\left(\frac{-4}{7}\right)}^n}\right)=-\frac{1}{7}$ và $\lim {\left(\frac{5}{7}\right)}^n=0$

Do đó $\lim u_n=0$

Chọn đáp án D.

Ta có: $u_n=\frac{n+\sqrt{n^2+1}}{n{.3}^n}=\frac{\frac{n+\sqrt{n^2+1}}{n}}{\frac{n{.3}^n}{n}}=\frac{1+\sqrt{1+\frac{1}{n^2}}}{3^n}=\frac{1}{3^n}\left(1+\sqrt{1+\frac{1}{n^2}}\right)$

Vì $\lim \frac{1}{n^2}=0$ nên $\lim \left(1+\sqrt{1+\frac{1}{n^n}}\right)=2$ và $\lim \frac{1}{3^n}=0.$

Do đó $\lim u_n=0$

Chọn đáp án D

Ta có:

$$\begin{gathered} \begin{array}{l}{u}_n=\sqrt[3]{n+2}-\sqrt[3]{n}=\frac{n+2-n}{{\left(\sqrt[3]{n+2}\right)}^2+\sqrt[3]{n+2}.\sqrt[3]{n}+{\left(\sqrt[3]{n}\right)}^2}\\ =\frac{2}{{\left(\sqrt[3]{n\left(1+\frac{2}{n}\right)}\right)}^2+\sqrt[3]{n\left(1+\frac{2}{n}\right)}.\sqrt[3]{n}+{\left(\sqrt[3]{n}\right)}^2}\end{array} \\ =\frac{2}{{\left(\sqrt[3]{n}\right)}^2\left[{\left(\sqrt[3]{1+\frac{2}{n}}\right)}^2+\sqrt[3]{1+\frac{2}{n}}+1\right]} \end{gathered}$$

Vì $\lim \frac{2}{3{\left(\sqrt[3]{n}\right)}^2}=0$ và $\lim \frac{1}{{\left(\sqrt[3]{1+\frac{2}{n}}\right)}^2+\sqrt[3]{1+\frac{2}{n}}+1}=\frac{1}{3}$

Do đó $\lim u_n=0$

Chọn đáp án B.

Xét tử số : $\sqrt{4n^2+1}-2n=\frac{4n^2+1-4n^2}{\sqrt{4n^2+1}+2n}=\frac{1}{\sqrt{4n^2+1}+2n}$

Xét mẫu số: $\frac{1}{\sqrt{n^2+4n+1}-n}=\frac{\sqrt{n^2+4n+1}+\text{}n}{n^2+4n+1-n^2}=\frac{\sqrt{n^2+4n+1}+n}{4n+1}$

$$\begin{gathered} \Rightarrow lim{u}_n=\lim \frac{\sqrt{n^2+4n+1}+n}{\left(\sqrt{4n^2+1}+2n\right)\left(2n+1\right)} \\ =\lim \frac{\sqrt{n^2\left(1+\frac{4}{n}+\frac{1}{n^2}\right)}+n}{\left(\sqrt{n^2\left(4+\frac{1}{n^2}\right)}+2n\right)\left(2n+1\right)} \\ =\lim \frac{n\left[\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1\right]}{n\left(\sqrt{4+\frac{1}{n^2}}+2\right)n\left(2+\frac{1}{n}\right)}=\lim \frac{\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1}{n\left(\sqrt{4+\frac{1}{n^2}}+2\right)\left(2+\frac{1}{n}\right)} \\ =\lim \frac{2}{n\left(2+2\right)2}=\lim \frac{1}{4n}=0. \end{gathered}$$ 

Do đó $\lim u_n=0$

Chọn đáp án C

* Xét tử số: Ta thấy 1, 2, 3, 4, ...,  n là một dãy số thuộc cấp số cộng có n số hạng với

$u_1 = 1$; d= 1 .

Tổng n số hạng của cấp số cộng: $S_n=\frac{\left(u_1+u_n\right)n}{2}=\frac{\left(1+n\right)n}{2}.$

* Xét mẫu số: Ta thấy $1,3,3^2,3^3,\mathrm{...},3^n$ là một dãy số thuộc cấp số nhân có n + 1 số hạng với $u_1 = 1$ ; q = 3

Tổng (n+ 1) số hạng của cấp số nhân: $S_{n+1}=u_1.\frac{1-q^{n+1}}{1-q}=\frac{1-3^{n+1}}{1-3}=\frac{3^{n+1}-1}{2}.$

$\Rightarrow u_n=\frac{n}{3^{n+1}-1}=\frac{n}{{3.3}^n-1}$

Bằng quy nạp ta luôn có $n<2^n,\forall n\in {\mathbb{N}}^{*}$ và $3^n>1,\forall n\in {\mathbb{N}}^{*}$

$\Rightarrow \left|u_n\right|=\left|\frac{n}{{3.3}^n-1}\right|<\frac{n}{3^n}<\frac{2^n}{3^n}={\left(\frac{2}{3}\right)}^n$

Vì $\lim {\left(\frac{2}{3}\right)}^n=0$ nên $\lim u_n=0.$

Chọn đáp án A

Ta có $\lim \frac{1-2n}{5n+5n^2}=\lim \frac{n^2\left(\frac{1}{n^2}-\frac{2}{n}\right)}{n^2\left(\frac{5}{n}+5\right)}=\lim \frac{\frac{1}{n^2}-\frac{2}{n}}{\frac{5}{n}+5}=\frac{0}{5}=0$.

Chọn đáp án D

Ta có $\left|\frac{\sin \left(n!\right)}{n^2+1}\right|\le \frac{1}{n^2+1}$ mà $\lim \frac{1}{n^2+1}=0$ nên $\lim \frac{\sin \left(n!\right)}{n^2+1}=0$

Chọn đáp án A.

Ta có $0<u_n=\frac{\sqrt{5^n}}{3^n+1}<\frac{{\left(\sqrt{5}\right)}^n}{3^n}={\left(\frac{\sqrt{5}}{3}\right)}^n;\forall n$ 

Lại có $\lim {\left(\frac{\sqrt{5}}{3}\right)}^n=0$ nên theo định lý kẹp có $\lim \frac{\sqrt{5^n}}{3^n+1}=0$

Chọn đáp án C

$$\begin{gathered} \lim {\left(2n+1\right)}^2\left(\frac{3}{n^2+2n}-\frac{1}{n^2+3n-1}\right) \\ =\lim \frac{{\left(2n+1\right)}^2\left(2n^2+7n-3\right)}{\left(n^2+2n\right)\left(n^2+3n-1\right)} \\ =\lim \frac{n^2{\left(2+\frac{1}{n}\right)}^2.n^2\left(2+\frac{7}{n}-\frac{3}{n^2}\right)}{n^2\left(1+\frac{2}{n}\right).n^2\left(1+\frac{3}{n}-\frac{1}{n^2}\right)} \\ =\lim \frac{{\left(2+\frac{1}{n}\right)}^2\left(2+\frac{7}{n}-\frac{3}{n^2}\right)}{\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}-\frac{1}{n^2}\right)}=\frac{2^2.2}{1.1}=8 \end{gathered}$$

Chọn đáp án D

$$\begin{gathered} \lim \frac{\sqrt{2n+2}-\sqrt{n}}{\sqrt{n}}=\lim \frac{\sqrt{n}.\sqrt{2+\frac{2}{n}}-\sqrt{n}}{\sqrt{n}} \\ =\lim \frac{\sqrt{2+\frac{2}{n}}-1}{1}=\sqrt{2}-1 \end{gathered}$$

Chọn đáp án A

$\lim \frac{\sqrt{2n+1}}{\sqrt{n}+1}=\lim \frac{\sqrt{n}.\sqrt{2+\frac{1}{n}}}{\sqrt{n}.\left(1+\sqrt{\frac{1}{n}}\right)}=\lim \frac{\sqrt{2+\frac{1}{n}}}{1+\sqrt{\frac{1}{n}}}=\sqrt{2}$

Chọn đáp án D

$$\begin{gathered} \lim \frac{4^{n+1}+6^{n+2}}{5^n+8^n}=\lim \frac{{4.4}^n+36.6^n}{5^n+8^n} \\ =\lim \frac{4.{\left(\frac{4}{8}\right)}^n+36.{\left(\frac{6}{8}\right)}^n}{{\left(\frac{5}{8}\right)}^n+1}=\frac{4.0+\text{}36.0}{0+\text{}1}=0 \end{gathered}$$

Chọn đáp án A

$$\begin{gathered} \lim \frac{2n^4-3n^3+2}{n^3+2}=\lim \frac{n^4\left(2-\frac{3}{n}+\frac{2}{n^4}\right)}{n^3\left(1+\frac{2}{n^3}\right)} \\ =\lim \left[n.\frac{2-\frac{3}{n}+\frac{2}{n^4}}{1+\frac{2}{n^3}}\right]=+\infty \end{gathered}$$

Vì $\lim n=+\infty ; \lim \frac{2-\frac{3}{n}+\frac{2}{n^4}}{1+\frac{2}{n^3}}=2$

$$\begin{gathered} \underset{x\to 3^{-}}{\lim }\frac{\left|x-3\right|}{5x-15}=\underset{x\to 3^{-}}{\lim }\frac{-x+3}{5x-15} \\ =\underset{x\to 3^{-}}{\lim }\frac{-(x-3)}{5(x-3)}=\underset{x\to 3^{-}}{\lim }\frac{-1}{5}=\frac{-1}{5} \end{gathered}$$

Chọn đáp án B

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\frac{2x+\sqrt{x}}{x-\sqrt{x}}=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)} \\ =\underset{x\to 0^{+}}{\lim }\frac{2\sqrt{x}+1}{\sqrt{x}-1}=\frac{1}{-1}=-1 \end{gathered}$$

Chọn đáp án D

Ta có:

$$\begin{gathered} \lim u_n=\lim \left(n-1\right)\sqrt{\frac{2n+2}{n^4+n^2-1}}=\lim \sqrt{\frac{{\left(n-1\right)}^2\left(2n+2\right)}{n^4+n^2-1}} \\ =\lim u_n=\lim \left(n-1\right)\sqrt{\frac{2n+2}{n^4+n^2-1}} \\ =\lim \sqrt{\frac{\frac{2}{n}-\frac{2}{n^2}-\frac{2}{n^3}+\frac{2}{n^4}}{1+\frac{1}{n^2}-\frac{1}{n^4}}}=0. \end{gathered}$$

Chọn đáp án B

$$\begin{gathered} L=\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}-\frac{1}{x^2-4}\right) \\ =\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}-\frac{1}{\left(x-2\right)\left(x+2\right)}\right) \\ =\underset{x\to 2^{-}}{\lim }\frac{x+2-1}{\left(x-2\right)\left(x+2\right)}=\underset{x\to 2^{-}}{\lim }\frac{x+1}{\left(x-2\right)\left(x+2\right)} \end{gathered}$$

Ta có 

$$\begin{gathered} x\to 2^{-}\Rightarrow x<2\Rightarrow x-2<0 \\ \Rightarrow \underset{x\to 2^{-}}{\lim }\left(x-2\right)=0,\underset{x\to 2^{-}}{\lim }\left(x+1\right)=3,\underset{x\to 2^{-}}{\lim }\left(x+2\right)=4 \end{gathered}$$

 Kết luận $L=-\infty$

Chọn đáp án B

$L=\underset{x\to -3^{+}}{\lim }\frac{2x^2+5x-3}{{\left(x+3\right)}^2}=\underset{x\to -3^{+}}{\lim }\frac{\left(2x-1\right)\left(x+3\right)}{{\left(x+3\right)}^2}=\underset{x\to -3^{+}}{\lim }\frac{2x-1}{x+3}$

Ta có $x\to -3^{+}\Rightarrow x>-3\Rightarrow x+3>0\Rightarrow \underset{x\to -3^{+}}{\lim }\left(x+3\right)=0$, và $\underset{x\to -3^{+}}{\lim }\left(2x-1\right)=-7.$

Kết luận $L=-\infty$

Chọn đáp án B.

$$\begin{gathered} \underset{x\to -\infty }{\lim }\frac{2\left|x\right|+3}{\sqrt{x^2+x+5}}=\underset{x\to -\infty }{\lim }\frac{-2x+3}{-x\sqrt{1+\frac{1}{x}+\frac{5}{x^2}}} \\ =\underset{x\to -\infty }{\lim }\frac{-2+\frac{3}{x}}{-\sqrt{1+\frac{1}{x}+\frac{5}{x^2}}}=2 \end{gathered}$$

Chọn đáp án C

$$\begin{gathered} \underset{x\to -\infty }{\lim }\frac{\sqrt{x^6+2}}{3x^3-1}=\underset{x\to -\infty }{\lim }\frac{\sqrt{x^6\left(1+\frac{2}{x^6}\right)}}{x^3\left(3-\frac{1}{x^3}\right)} \\ =\underset{x\to -\infty }{\lim }\frac{-x^3\sqrt{1+\frac{2}{x^3}}}{x^3\left(3-\frac{1}{x^3}\right)}=\underset{x\to -\infty }{\lim }\frac{-\sqrt{\left(1+\frac{2}{x^3}\right)}}{3-\frac{1}{x^3}}=\frac{-1}{3} \end{gathered}$$

Chọn đáp án A

Ta có:

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(x+m\right)=m; \underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\left(x^2+1\right)=1$

Hàm số có giới hạn tại $x=0\iff \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)\iff m=1$

Chọn đáp án D

Đặt 

$$\begin{gathered} u_n=\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{....}+\frac{1}{n\left(n+1\right)} \\ =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\mathrm{...}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1} \end{gathered}$$

$\Rightarrow \lim u_n=\lim \left(1-\frac{1}{n+1}\right)=1$

Chọn đáp án B

Tại điểm x= -1 ta có:

$$\begin{gathered} \underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{-}}{\lim }\left(3x+b\right)=-3+\text{}b=f(-1) \\ \underset{x\to {\left(-1\right)}^{+}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{+}}{\lim }\left(x+a\right)=-1+a \end{gathered}$$

Hàm số có giới hạn tại x= -1 khi và chỉ khi $\underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{+}}{\lim }f\left(x\right)$

Điều này tương đương với $-3+b=-1+a\Rightarrow a-b=-2$

Chọn đáp án B.

Ta có:

$\left\{\begin{array}{l}\underset{x\to -1^{-}}{\lim }h\left(x\right)=\underset{x\to -1^{-}}{\lim }\frac{x^3+1}{x+1}=\underset{x\to -1^{-}}{\lim }\left(x^2-x+1\right)=3\\ \underset{x\to -1^{+}}{\lim }h\left(x\right)=\underset{x\to -1^{+}}{\lim }\left(m{x}^2-x+m^2\right)=m+1+\text{}{m}^2\end{array}\right.$

Hàm số có giới hạn tại x= -1 khi và chỉ khi $\underset{x\to -1^{-}}{\lim }h\left(x\right)=\underset{x\to -1^{+}}{\lim }h\left(x\right)$

$3=m+1+\text{\hspace{0.17em}}{m}^2\iff m^2+m-2=0\iff \left[\begin{array}{l}m=1\\ m=-2\end{array}\right.$

Chọn đáp án C

Ta có :

$\left\{\begin{array}{l}\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{-}}{\lim }m=m\\ \underset{x\to 3^{+}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }\frac{3-x}{\sqrt{x+1}-2}=\underset{x\to 3^{+}}{\lim }\frac{(3-x).\text{}(\sqrt{x+1}+2)}{x-3}=\underset{x\to 3^{+}}{\lim }-(\sqrt{x+1}+2)=-4\end{array}\right.$

Vậy để tồn tại $\underset{x\to 3}{\lim }f\left(x\right)$ khi và chỉ khi :

$\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }f\left(x\right)\iff m=-4$ 

Chọn đáp án C

Với x= 1 ta có :$f\left(1\right)=k^2$

Với $x\ne 1$ ta có :

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(x^2+3\right)=4; \underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }{\left(x+1\right)}^2=4$ suy ra $\underset{x\to 1}{\lim }f\left(x\right)=4$

Vậy để hàm số gián đoạn tại x= 1 khi $\underset{x\to 1}{\lim }f\left(x\right)\ne k^2\iff k^2\ne 4\iff k\ne \pm 2$

Chọn đáp án A

Ta có : $f\left(-1\right)=1$ và $\underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{-}}{\lim }(2x+\text{}3)=2.(-1)\text{}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3=1$

$$\begin{gathered} \underset{x\to {\left(-1\right)}^{+}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{+}}{\lim }\frac{x+\sqrt{x+2}}{x+1} \\ =\underset{x\to {\left(-1\right)}^{+}}{\lim }\frac{x^2-x-2}{\left(x+1\right)\left(x-\sqrt{x+2}\right)} \\ =\underset{x\to {\left(-1\right)}^{+}}{\lim }\frac{x-2}{x-\sqrt{x+2}}=\frac{3}{2} \end{gathered}$$

Suy ra : $\underset{x\to {\left(-1\right)}^{+}}{\lim }f\left(x\right)\ne \underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)$

Vậy hàm số gián đoạn tại x= -1.

Chọn đáp án C

$\underset{x\to 2}{\lim }\sqrt{\frac{4x^2-3x}{\left(2x-1\right)\left(x^3-2\right)}}=\sqrt{\frac{{4.2}^2-3.2}{\left(2.2-1\right)\left(2^3-2\right)}}=\sqrt{\frac{10}{18}}=\frac{\sqrt{5}}{3}.$

Chọn đáp án D

Ta có $4x^5-3x^3+x+1=x^5\left(4-\frac{3}{x^2}+\frac{1}{x^4}+\frac{1}{x^5}\right)$

Vì $\underset{x\to -\infty }{\lim }{x}^5=-\infty ;\underset{x\to -\infty }{\lim }\left(4-\frac{3}{x^2}+\frac{1}{x^4}+\frac{1}{x^5}\right)=4>0$

Vậy: $\underset{x\to -\infty }{\lim }\left(4x^5-3x^3+x+1\right)=-\infty$

Chọn đáp án D

$A=\underset{x\to 2}{\lim }\frac{3x+m}{x+2}=\frac{6+m}{4},$

Mà $A=5\Rightarrow \frac{6+m}{4}=5\iff 6+\text{}m=20\iff m=14$

Chọn đáp án B

$\underset{x\to -1}{\lim }\frac{x^3+x^2}{{\left(x+1\right)}^3}=\underset{x\to -1}{\lim }\frac{x^2\left(x+1\right)}{{\left(x+1\right)}^3}=\underset{x\to -1}{\lim }\frac{x^2}{{\left(x+1\right)}^2}.$

Vì: $\underset{x\to -1}{\lim }{x}^2=1>0;\underset{x\to -1}{\lim }{\left(x+1\right)}^2=0$ và ${\left(x+1\right)}^2>0,\forall x\ne -1$

nên $\underset{x\to -1}{\lim }\frac{x^3+x^2}{{\left(x+1\right)}^3}=+\infty$

Chọn đáp án D

Ta có:

$\begin{array}{l}\underset{x\to -\infty }{\lim }\left(\sqrt{5x^2+2x}+x\sqrt{5}\right)=\underset{x\to -\infty }{\lim }\frac{2x}{\sqrt{5x^2+2x}-x\sqrt{5}}\\ =\underset{x\to -\infty }{\lim }\frac{2}{-\sqrt{5+\text{\hspace{0.17em}}\frac{2}{x}}-\sqrt{5}}=-\frac{\sqrt{5}}{5}.\end{array}$

Chọn đáp án B

$$\begin{gathered} \underset{x\to -\infty }{\lim }\left(\sqrt{x^2+ax+5}+x\right)=\underset{x\to -\infty }{\lim }\frac{a.x+5}{\sqrt{x^2+ax+5}-x} \\ =\underset{x\to -\infty }{\lim }\frac{a+\frac{5}{x}}{-\sqrt{1+\frac{a}{x}\text{}+\text{}\frac{5}{x^2}}-1}=-\frac{a}{2} \end{gathered}$$