Question

limit x tends to 3 logx - log3/x-3​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 3}\rm \frac{logx - log3}{x - 3}$

If we substitute directly x = 3, we get

$\rm \:  =  \: \dfrac{log3 - log3}{3 - 3}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 3}\rm \frac{logx - log3}{x - 3}$

To evaluate this limit, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = 3 + h, \: \: as \: x \: \to \: 3 \: \: so \: \: h \: \to \: 0}$

So, above expression can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log(3 + h) - log3}{3 + h - 3}$

We know,

$\boxed{\tt{ logx - logy = log \frac{x}{y} \: }}$

So, using this, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[\dfrac{3 + h}{3} \bigg]}{h}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[1 + \dfrac{h}{3} \bigg]}{h}$

can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[1 + \dfrac{h}{3} \bigg]}{\dfrac{h}{3} \times 3 }$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[1 + \dfrac{h}{3} \bigg]}{\dfrac{h}{3}} \times \displaystyle\lim_{h \to 0}\rm \frac{1}{3}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: 1 \times \dfrac{1}{3}$

$\rm \:  =  \: \dfrac{1}{3}$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle\lim_{x \to 3}\rm \frac{logx - log3}{x - 3} = \frac{1}{3}}}$

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ADDITIONAL INFORMATION

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

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

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

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

Answer 2

Answer:

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 3}\rm \frac{logx - log3}{x - 3}$

If we substitute directly x = 3, we get

$\rm \:  =  \: \dfrac{log3 - log3}{3 - 3}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 3}\rm \frac{logx - log3}{x - 3}$

To evaluate this limit, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = 3 + h, \: \: as \: x \: \to \: 3 \: \: so \: \: h \: \to \: 0}$

So, above expression can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log(3 + h) - log3}{3 + h - 3}$

We know,

$\boxed{\tt{ logx - logy = log \frac{x}{y} \: }}$

So, using this, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[\dfrac{3 + h}{3} \bigg]}{h}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[1 + \dfrac{h}{3} \bigg]}{h}$

can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[1 + \dfrac{h}{3} \bigg]}{\dfrac{h}{3} \times 3 }$

$\rm \:  =  \: \displaystyle\lim_{h \to 0}\rm \frac{log\bigg[1 + \dfrac{h}{3} \bigg]}{\dfrac{h}{3}} \times \displaystyle\lim_{h \to 0}\rm \frac{1}{3}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: 1 \times \dfrac{1}{3}$

$\rm \:  =  \: \dfrac{1}{3}$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle\lim_{x \to 3}\rm \frac{logx - log3}{x - 3} = \frac{1}{3}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

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

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

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

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