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# Question:Examine the following functions for continuity.(a) $f(x) = x - 5$(b) $f(x) = \dfrac{1}{x - 5} \ x \neq 5$(c) $f(x) = \dfrac{x^2 - 25}{x+5} \ x \neq -5$(d) $f(x) = |x - 5|$

Last updated date: 19th Sep 2024
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Hint: A function is continuous at a point if the one-sided limits at that point exist and are equal to the function's value at that point.

Solution:

(i) The function is a first-degree polynomial. All polynomial functions are continuous over the entire real line, $\mathbb{R}$. Therefore, $f$ is continuous for all $x \in \mathbb{R}$.

(ii) For the function $f(x) = \dfrac{1}{{x-5}}$, where $x \neq 5$:

For any $k \neq 5$, we have:

$\lim_{{x \to k}} f(x) = \lim_{{x \to k}} \dfrac{1}{{x-5}} = \dfrac{1}{{k-5}}$​

Also, $f(k) = \dfrac{1}{{k-5}}$ since $k \neq 5$.

Therefore, the function is continuous at every point in its domain.

(iii) For the function $f(x) = \dfrac{{x+5}}{{x^2 - 25}}$, where $x \neq -5$:

For any real $x \neq -5$, we get:

$\lim_{{x \to c}} f(x) = \lim_{{x \to c}} \dfrac{{x+5}}{{(x+5)(x-5)}} = x-5$

Thus, the function is continuous at every point in its domain.

(iv) For the function $f(x) = |x - 5|$:

This function can be expressed piecewise as:

$f(x) = \begin{cases} 5 - x & \text{if } x < 5 \\ x - 5 & \text{if } x \geq 5 \end{cases}$​

For $c < 5$, $f(c) = 5 - c$ and $\lim_{{x \to c}} f(x) = 5 - c$, proving continuity for $x < 5$.

At $c = 5$, $f(c) = 0$ and both left and right limits as $x \to 5$ also equal to 0. Therefore, $f$ is continuous at $x = 5$.

For $c > 5$, $f(c) = c - 5$ and $\lim_{{x \to c}} f(x) = c - 5$, proving continuity for $x > 5$.

Thus, the function is continuous for all real numbers.

In summary, each function given has been proven to be continuous over its domain.