Let's evaluate both statements:
Statement 1: We are given that \[ \lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} \] Substituting \( x = 1 \) into the expression, we get \[ \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a} = \frac{a + b + c}{c + b + a} = 1 \] Thus, statement 1 is true.
Statement 2: We are given that \[ \lim_{x \to -2} \frac{1}{x + 2} \] As \( x \to -2 \), the denominator approaches 0, making the limit undefined. Therefore, statement 2 is false.
The correct answer is (B) : Only statement 1 is true
Prove that the function \( f(x) = |x| \) is continuous at \( x = 0 \) but not differentiable.
\[ f(x) = \begin{cases} x^2 + 2, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \]
is not continuous at \( x = 0 \).Is the function \( f(x) \) defined by
\[ f(x) = \begin{cases} x + 5, & \text{if } x \leq 1 \\ x - 5, & \text{if } x > 1 \end{cases} \]
continuous at \( x = 1 \)?