Question

$\boxed{ \huge\bold \red { { \displaystyle \lim_{n \to \infty} } \frac{4 {n}^{2} }{ {n}^{2} + 10000 \: n } }}$


❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​​​​​​​​​​​​​​​​​​

Answer 1

$\lim_{n \to \infty} \frac{4 n^{2}}{n^{2} + 10000 n}$

Apply the constant multiple rule

$\bold{\lim_{n \to \infty} \: c f{\left(n \right)} = c \lim_{n \to \infty} \: f{\left(n \right)} \: with \: c=4 \: and f{\left(n \right)} = \frac{n}{n + 10000}:}$

$\bold{\color{red}{\lim_{n \to \infty} \frac{4 n^{2}}{n^{2} + 10000 n}} = \color{red}{\left(4 \lim_{n \to \infty} \frac{n}{n + 10000}\right)}}$

Multiply and divide by n:

$\bold{ \color{red}{4\lim_{n \to \infty} \frac{n}{n + 10000}} = 4 \color{red}{\lim_{n \to \infty} \frac{n}{n \frac{n + 10000}{n}}}}$

Divide:

$\bold{ \color{red}{4\lim_{n \to \infty} \frac{n}{n \frac{n + 10000}{n}}} = 4 \color{red}{\lim_{n \to \infty} \frac{1}{1 + \frac{10000}{n}}}}$

The limit of the quotient is the quotient of limits:

$4 \color{red}{\lim_{n \to \infty} \frac{1}{1 + \frac{10000}{n}}} = 4 \color{red}{\frac{\lim_{n \to \infty} 1}{\lim_{n \to \infty}\left(1 + \frac{10000}{n}\right)}}$

The limit of a constant is equal to the constant:

$\bold{ \blue{\frac{4 \color{blue}{\lim_{n \to \infty} 1}}{\lim_{n \to \infty}\left(1 + \frac{10000}{n}\right)} = \frac{4 \color{red}{1}}{\lim_{n \to \infty}\left(1 + \frac{10000}{n}\right)}}}$

The limit of a sum/difference is the sum/difference of limits:

$\bold \pink{4 \color{red}{\lim_{n \to \infty}\left(1 + \frac{10000}{n}\right)}^{-1} = 4 \color{red}{\left(\lim_{n \to \infty} 1 + \lim_{n \to \infty} \frac{10000}{n}\right)}^{-1}}$

The limit of a constant is equal to the constant:

${\bold \pink{4 \left(\lim_{n \to \infty} \frac{10000}{n} + \color{brown}{\lim_{n \to \infty} 1}\right)^{-1} = 4 \left(\lim_{n \to \infty} \frac{10000}{n} + \color{red}{1}\right)^{-1}}}$

Apply the constant multiple rule:

${\bold \blue{\lim_{n \to \infty} c f{\left(n \right)} = c \lim_{n \to \infty} f{\left(n \right)} \: with \: c=10000\: and \: f{\left(n \right)} = \frac{1}{n}}}$

$\bold{{\color{brown}4\left(1 + \color{red}{\lim_{n \to \infty} \frac{10000}{n}}\right)^{-1} = 4 \left(1 + \color{red}{\left(10000 \lim_{n \to \infty} \frac{1}{n}\right)}\right)^{-1}}}$

The limit of a quotient is the quotient of limits:

${ \pink{4 \left(1 + 10000 \color{red}{\lim_{n \to \infty} \frac{1}{n}}\right)^{-1} = 4 \left(1 + 10000 \color{red}{\frac{\lim_{n \to \infty} 1}{\lim_{n \to \infty} n}}\right)^{-1}}}$

The limit of a constant is equal to the constant:

${4 \left(1 + \frac{10000 \color{red}{\lim_{n \to \infty} 1}}{\lim_{n \to \infty} n}\right)^{-1} = 4 \left(1 + \frac{10000 \color{red}{1}}{\lim_{n \to \infty} n}\right)^{-1}}$

Constant divided by a very big number equals 0:

${4 \left(1 + 10000 \color{red}{1 \frac{1}{\lim_{n \to \infty} n}}\right)^{-1} = 4 \left(1 + 10000 \color{red}{\left(0\right)}\right)^{-1}}$

Therefore,

$\bold{\lim_{n \to \infty} \frac{4 n^{2}}{n^{2} + 10000 n} = 4}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\red{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{ 4{n}^{2} }{ {n}^{2} + 10000n}}$

If we substitute directly, the value of x, we get

$\rm \:  =  \:\dfrac{ \infty }{ \infty }$

which is indeterminant form.

So,

$\red{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{ 4{n}^{2} }{ {n}^{2} + 10000n}}$

On dividing numerator and denominator by n^2, we get

$\rm \:  =  \:\displaystyle\lim_{n \to \infty } \frac{\dfrac{ {4n}^{2} }{ {n}^{2} } }{\dfrac{ {n}^{2} + 10000n}{ {n}^{2} } }$

$\rm \:  =  \:\displaystyle\lim_{n \to \infty } \frac{4}{\dfrac{ {n}^{2} }{ {n}^{2}} + \dfrac{10000n}{ {n}^{2} } }$

$\rm \:  =  \:\displaystyle\lim_{n \to \infty } \frac{4}{1 + \dfrac{10000}{ {n}} }$

$\rm \:  =  \:\dfrac{4}{1 + \dfrac{10000}{ \infty } }$

$\rm \:  =  \:\dfrac{4}{1 + 0}$

$\rm \:  =  \:4$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{n \to \infty} \frac{ 4{n}^{2} }{ {n}^{2} + 10000n}} = 4 \: }}$

Alternative Method :-

Given function is

$\red{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{ 4{n}^{2} }{ {n}^{2} + 10000n}}$

On applying L Hospital Rule, we get

$\rm \:  =  \:\displaystyle\lim_{n \to \infty } \frac{\dfrac{d}{dn} {4n}^{2} }{\dfrac{d}{dn}( {n}^{2} + 10000n) }$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

So, using this, we get

$\rm \:  =  \:\displaystyle\lim_{n \to \infty } \frac{8n}{2n + 10000}$

Again, applying L Hospital Rule, we get

$\rm \:  =  \:\displaystyle\lim_{n \to \infty } \frac{\dfrac{d}{dn}8n}{\dfrac{d}{dn}(2n + 10000)}$

$\rm \:  =  \:\displaystyle\lim_{n \to \infty } \frac{8}{2 + 0}$

$\rm \:  =  \:\dfrac{8}{2}$

$\rm \:  =  \:4$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{n \to \infty} \frac{ 4{n}^{2} }{ {n}^{2} + 10000n}} = 4 \: }}$

More to know

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0 } \frac{sinx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0 } \frac{tanx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0 } \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0 } \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0 } \frac{ {a}^{x} - 1}{x} = loga \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0 } \frac{ 1 - cosx}{x} = 0 \: }}$