Question

If we have a function as $f(x) = \dfrac{{{5^x} + {5^{ - x}} - 2}}{{{x^2}}}$ , for $x \ne 0$
$= k$, for x=0
Is continuous at x=0 find k.

Answer 1

A function f(x) is said to be continuous at x=a, if the given function is defined at x=a and if the value of the given function at x=a is equal to the limit of the given function at x=a i.e. $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$. If these conditions are not followed then we say that the given function is discontinuous at x=a.

Complete step-by-step solution:
Step 1: According to the question it is given that the given function f(x) is continuous at x=0, so we should have a finite value of the given function at x=0 and in the question, it is provided that
At x=0, $f(0) = k$
Since the given function f(x) is continuous at x=0 so we can say that the value of the given function at x=0 is equal to the limit of the given function at x=0 i.e.
$\mathop {\lim }\limits_{x \to 0} f(0) = f(0)$
Step 2: Now according to the continuity conditions we have,
$\mathop {\lim }\limits_{x \to 0} f(0) = f(0)$&$f(0) = k$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = f(0) = k$
According to the question,
$f(x) = \dfrac{{{5^x} + {5^{ - x}} - 2}}{{{x^2}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{5^x} + {5^{ - x}} - 2}}{{{x^2}}}$
By rearranging the terms we further proceed as,
$\mathop {\lim }\limits_{x \to 0} f(0) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{5^x} + \dfrac{1}{{{5^x}}} - 2}}{{{x^2}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{({5^x})}^2} + 1 - 2({5^x})}}{{{x^2}}}$
As we know from the concept of a quadratic equation that, ${a^2} - 2ab + {b^2} = {(a - b)^2}$ , so we have
$\mathop {\lim }\limits_{x \to 0} f(0) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{({5^x} - 1)}^2}}}{{{5^x} \cdot {x^2}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{{5^x} - 1}}{x}} \right)^2} \cdot \dfrac{1}{{{5^x}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{{5^x} - 1}}{x}} \right)^2} \times \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{5^x}}}$
Step 3: From the limit of exponential functions we know that
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x} = \ln a(a > 0)$
From the above expression, we can write
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{{5^x} - 1}}{x}} \right)^2} \times \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{5^x}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = {(\log 5)^2} \times \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{5^x}}}$
Taking$x \to 0$, we get
$\mathop {\lim }\limits_{x \to 0} f(0) = {(\log 5)^2} \times \dfrac{1}{{{5^0}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = {(\log 5)^2} \times 1$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f(0) = {(\log 5)^2}$
According to the question, $\mathop {\lim }\limits_{x \to 0} f(0) = f(0) = k$
$k = {(\log 5)^2}$
Hence, the value of k is equal to ${(\log 5)^2}$.

Note: A given function is said to be discontinuous when,

The limit does not exist, that is
$\mathop {\lim }\limits_{x \to {a^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {a^ + }} f(x)$
F(x) is not defined at x=0
$\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)$
Geometrically, the graph of the function will exhibit a break at x=a, if the function is discontinuous at x=a.