Examine whether the function $f$ given by $f(x)=x^{2}$ is continuous at $x=0$.

Step 1: Understand Continuity Condition

For a function $f(x)$ to be continuous at a point $x=a$, three conditions must be met: the function must be defined at $a$, the limit of the function as $x$ approaches $a$ must exist, and these two values must be equal. In this problem, we need to check continuity at $x=0$.

Step 2: Evaluate the function at $x=0$

First, we evaluate the function $f(x) = x^2$ at the given point $x=0$. Substituting $x=0$ into the function, we find that $f(0)$ is $0^2$, which equals $0$.

Step 3: Evaluate the limit as x approaches 0

Next, we need to find the limit of the function $f(x) = x^2$ as $x$ approaches $0$. Since $f(x)$ is a polynomial function, its limit as $x$ approaches any point can be found by direct substitution.

Step 4: Calculate the limit

By substituting $x=0$ into the expression $x^2$, we find that the limit of $f(x)$ as $x$ approaches $0$ is $0^2$, which is $0$.

Step 5: Compare function value and limit

We found that $f(0) = 0$ and $\lim_{x \to 0} f(x) = 0$. Since these two values are equal, the condition for continuity is satisfied.

More from this chapter

• 2. Consider the function f(x) = 3cos(e^(5x-3)) + (4)/(x) Find f'(x) [(d)/(dx)(cos x) = -sin x]
• **प्रश्नावली 5.2** प्रश्न 1 से 8 में x के सापेक्ष निम्नलिखित फलनों का अवकलन कीजिए: 1. sin(x² + 5) 2. cos(sin x) 3. sin(ax + b) 4. sec(tan(√x)) 5. (sin(ax + b))/
• 1. Prove that the function f(x) = 2x² - 3x + 2 is continuous at x = 1. 2. The function f(x) is defined as follows: f(x) = 2x - 3, & x < 2 x - 1, & x ≥ 2 Prove t
• Find dy/dx if y = (ln(x))^(cos(x)).
• Differentiate the following w.r.t. x. (i) cos^-1(sin x) (ii) tan^-1(sin x / (1 + cos x)) (iii) sin^-1(2^(x+1) / (1 + 4^x))
• If x = t² and y = t³, then (d² y)/(dx²) is equal to:
• Find (d y)/(d x), if x=a(θ+sin θ), y=a(1-cos θ).
• Differentiate the following w.r.t. x. (i) cos ^(-1)(sin x) (ii) tan ^(-1)((sin x)/(1+cos x)) (iii) sin ^(-1)((2^(x+1))/(1+4^x))

All Continuity and Differentiability solutions →