Question

Evaluate the given $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {x + 5} \right)}^5} - 1}}{x}$ .

Answer 1

This is the problem from limits and derivatives. We need to evaluate the limit here. But we will use one of the formulas or property of limits.
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n.{a^{n - 1}}$

Complete step-by-step answer:
Given that,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {x + 5} \right)}^5} - 1}}{x}$
Here let $x + 5 = y$
$\Rightarrow x = y - 5$
Now we will replace x by y-5 in the limits above
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {y - 5 + 5} \right)}^5} - 1}}{{y - 5}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( y \right)}^5} - 1}}{{y - 5}}$
But we can write $1 = {1^5}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( y \right)}^5} - {1^5}}}{{y - 5}}$
Now limit above is of the form
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^n} - {a^n}}}{{x - a}}$
So we can write it in the form $n.{a^{n - 1}}$

$$\Rightarrow 5 \times {1^{5 - 1}} \\\ \Rightarrow 5 \times {1^4} \\\ \Rightarrow 5 \times 1 \\\ \Rightarrow 5 \\\$$

Thus
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {x + 5} \right)}^5} - 1}}{x} = 5$
This is our correct answer.

Note: Sometimes students directly write the value of x as 0 and try to solve but that will lead to 0 only. So first simplify the equation and then solve.
Some formulas like:

$$\mathop {\lim }\limits_{x \to 0} {e^x} = 1 \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = 1 \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x} = {\log _e}a \\\$$