If $f$ is continuous at $5$ and $f(5)=2$ and $f(4)=3$, then $\lim_{x \to 2} f(4x^2-11)=2$
Is the above statement true or false with an explanation for the answer?
I'm not sure how to approach this.
If $f$ is continuous at $5$ and $f(5)=2$ and $f(4)=3$, then $\lim_{x \to 2} f(4x^2-11)=2$
Is the above statement true or false with an explanation for the answer?
I'm not sure how to approach this.
Theorem: if $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$, then the composite function $f(g(x))=fog(x)$ is continuous at $x_0$.
As you noted $f$ is continuous at $5$ and you know that polynomials are continuous at every real numbers especially at $x_0=2$. so $\lim_{x\to2}f(4x^2-11)=f(\lim_{x\to2}(4x^2-11))=f(4*2^2-11)=f(5)=2$
Since $f$ is continuous at $5$, and $5$ is the limit of $4x^2-11$ as $x$ approaches $2$, you make take the limit "inside the function" to see that the value of the limit is $f(5)=2$.