Question

limit x tends to a where x to the power -5 -a to the power -5 divided by x to the power -7 -a to the power -7
​

Answer 1

Given Question :-

Evaluate the following limit

$\rm \: \displaystyle\lim_{x \to a}\rm \: \dfrac{ {x}^{ - 5} - {a}^{ - 5} }{ {x}^{ - 7} - {a}^{ - 7}}$

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

Given expression is

$\rm \: \displaystyle\lim_{x \to a}\rm \: \dfrac{ {x}^{ - 5} - {a}^{ - 5} }{ {x}^{ - 7} - {a}^{ - 7}}$

If we substitute directly x = a, we get

$\rm \: = \: \dfrac{ {a}^{ - 5} - {a}^{ - 5}}{ {a}^{ - 7} - {a}^{ - 7}}$

$\rm \: = \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to a}\rm \: \dfrac{ {x}^{ - 5} - {a}^{ - 5} }{ {x}^{ - 7} - {a}^{ - 7}}$

Divide numerator and denominator by x - a, we get

$\rm \: = \: \displaystyle\lim_{x \to a}\rm \dfrac{ \dfrac{ {x}^{ - 5} - {a}^{ - 5}}{x - a} }{ \: \: \dfrac{ {x}^{ - 7} - {a}^{ - 7}}{x - a} \: \: }$

can be further rewritten as

$\rm \: = \: \dfrac{\displaystyle\lim_{x \to a}\rm \: \frac{ {x}^{ - 5} - {a}^{ - 5}}{x - a} }{\displaystyle\lim_{x \to a}\rm \: \frac{ {x}^{ - 7} - {a}^{ - 7}}{x - a} }$

We know,

$\boxed{\rm{  \:\displaystyle\lim_{x \to a}\rm \frac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} \: \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{ - 5 {a}^{ - 5 - 1} }{ - 7 {a}^{ - 7 - 1} }$

$\rm \: = \: \dfrac{5 {a}^{ - 6} }{7 {a}^{ - 8} }$

$\rm \: = \: \dfrac{5}{7} {a}^{2}$

Hence,

$\rm\implies \:\boxed{\rm{  \:\rm \: \displaystyle\lim_{x \to a}\rm \: \dfrac{ {x}^{ - 5} - {a}^{ - 5} }{ {x}^{ - 7} - {a}^{ - 7}} = \frac{5}{7} {a}^{2} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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

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

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

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

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