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

Hint: 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:
\[
  \mathop {\lim }\limits_{x \to 0} {e^x} = 1 \\
  \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = 1 \\
  \mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x} = {\log _e}a \\
\]