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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.

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The composition of continuous functions is continuous. So $\lim_{x\to a} f(g(x))=f(\lim_{u\to g(a)}u)=f(g(a))$ when $g$ is continuous at $a$ and $f$ is continuous at $g(a)$. Apply this with $g(x)=4x^2-11$. –  anon Aug 23 '12 at 18:57
    
The fact that $f(4)=3$ is totally irrelevant and is just meant to confuse and distract. –  André Nicolas Aug 23 '12 at 19:04
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2 Answers

up vote 3 down vote accepted

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 $$\lim_{x\to2}f(4x^2-11)=f(\lim_{x\to2}(4x^2-11))=f(4*2^2-11)=f(5)=2$$

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Sorry. I'm not trying to copy you. I was typing (while trying to talk on the phone) as you were answering, I guess. –  Chris Leary Aug 23 '12 at 19:06
    
@ChrisLeary: It is Ok. :) Anyway, anon gave him the final answer at above comment first. –  B. S. Aug 23 '12 at 19:15
    
Thank you both for your responses. Very helpful! –  AKrush95 Aug 23 '12 at 19:24
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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 $f(5)=2$.

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