Answer to Question #131641 in Calculus for Temare

Question

Prove that limit x approach to 1 (28/(4x+3))=4

Answer 1

We can prove it by epsilon and delta definition.

By definition, $\lim_{x\to a} f(x)=b$ if and only if for every $ε>0$ there is a $δ>0$ such that: for all $x$, if $0<|x−a|<δ$, then $|f(x)−b|<ε$.

As we want to show that $\lim_{x\to 1} \frac{28}{4x+3}=4$, we have to make $|\frac{28}{4x+3}-4|<ε$ and we control $|x-1|<δ$.

$|\frac{28}{4x+3}-4|=|\frac{28-4(4x+3)}{4}|=|\frac{28-16x-12}{4}|=|\frac{16-16x}{4}|=|\frac{16(1-x)}{4}|=|4(1-x)|$.

In order to make $|4(x−1)|<ε$, it suffices to make $|x−1|<\frac{ε}{4}$.

Proof. Let $ε>0$ be given, choose $δ=\frac{ε}{4}$ and assume that $0 < | x – 1| <δ$.

We have: if $0 < | x – 1| <δ$ then

$|\frac{28}{4x+3}-4|=|4(1-x)|=4|1-x|=4|x-1|<4δ=4\frac{ε}{4}=ε$.

Thus, we conclude that,  $|\frac{28}{4x+3}-4|<ε$.

We have shown that for any positive ε, there is a positive δ such that for all $x$, if $0<|x−1|<δ$, then $|\frac{28}{4x+3}-4|<ε$.

So, by the definition of limit, we have $\lim_{x\to 1} \frac{28}{4x+3}=4$.

The other way to prove it:

As the function $f(x)=\frac{28}{4x+3}$ is continuous at point $x=1$, then $\lim_{x\to 1} f(x)=f(1)=\frac{28}{4 \cdot 1+3}=\frac{28}{7}=4.$