Question

1. evaluate the limit, if it exists. $lim_{x

ightarrow4}\frac{x^{2}-4x}{x^{2}-3x - 4}$

1. if $2xleq g(x)leq x^{4}-x^{2}+2$ for all $x$, evaluate $lim_{x

ightarrow1}g(x)$.

Explanation:

5.

Step1: Factor the numerator and denominator

The numerator $x^{2}-4x=x(x - 4)$. The denominator $x^{2}-3x - 4=(x - 4)(x+1)$. So, $\lim_{x
ightarrow4}\frac{x^{2}-4x}{x^{2}-3x - 4}=\lim_{x
ightarrow4}\frac{x(x - 4)}{(x - 4)(x + 1)}$.

Step2: Cancel out the common factor

Since $x
eq4$ when taking the limit, we can cancel out the $(x - 4)$ terms. We get $\lim_{x
ightarrow4}\frac{x}{x + 1}$.

Step3: Substitute $x = 4$

Substitute $x=4$ into $\frac{x}{x + 1}$, we have $\frac{4}{4+1}=\frac{4}{5}$.

Step1: Find the limits of the bounding - functions

First, find $\lim_{x
ightarrow1}2x$. Substitute $x = 1$ into $2x$, we get $\lim_{x
ightarrow1}2x=2\times1 = 2$.
Next, find $\lim_{x
ightarrow1}(x^{4}-x^{2}+2)$. Substitute $x = 1$ into $x^{4}-x^{2}+2$, we have $1^{4}-1^{2}+2=1 - 1+2=2$.

Step2: Apply the Squeeze Theorem

Since $2x\leq g(x)\leq x^{4}-x^{2}+2$ for all $x$ and $\lim_{x
ightarrow1}2x=\lim_{x
ightarrow1}(x^{4}-x^{2}+2)=2$, by the Squeeze Theorem, $\lim_{x
ightarrow1}g(x)=2$.

Answer:

$\frac{4}{5}$

6.