Question

1. find the limit.

a. $lim_{x
ightarrow - 1}(2x^{2}-3x)$
b. $lim_{x
ightarrow 3}\frac{x^{2}-2x - 3}{x^{2}-5x + 6}$
c. $lim_{\theta
ightarrow 0}\frac{3\theta+cos\theta}{5sin\theta + 1}$
d. $lim_{x
ightarrowinfty}(\frac{-3}{x^{4}}+1)$
e. $lim_{x
ightarrow 5^{+}}\frac{-1}{(x - 5)^{2}}$
f. $lim\frac{-1}{}
g. $lim_{x
ightarrow 5^{+}}\frac{1}{x - 5}$
h. $lim_{x
ightarrow 5^{-}}\frac{1}{x - 5}$
i. $lim\frac{cos\theta}{}$

Explanation:

Step1: Substitute the value for a

For $\lim_{x
ightarrow - 1}(2x^{2}-3x)$, substitute $x = - 1$ into the function.
$2(-1)^{2}-3(-1)$

Step2: Calculate the result for a

$2\times1 + 3=2 + 3=5$

Step3: Factor the numerator and denominator for b

For $\lim_{x
ightarrow3}\frac{x^{2}-2x - 3}{x^{2}-5x + 6}$, factor $x^{2}-2x - 3=(x - 3)(x+1)$ and $x^{2}-5x + 6=(x - 3)(x - 2)$.
$\lim_{x
ightarrow3}\frac{(x - 3)(x + 1)}{(x - 3)(x - 2)}$

Step4: Simplify and find the limit for b

Cancel out the common factor $(x - 3)$ and then substitute $x = 3$.
$\lim_{x
ightarrow3}\frac{x + 1}{x - 2}=\frac{3+1}{3 - 2}=4$

Step5: Substitute the value for c

For $\lim_{\theta
ightarrow0}\frac{3\theta+\cos\theta}{5\sin\theta + 1}$, substitute $\theta=0$.
$\frac{3\times0+\cos(0)}{5\sin(0)+1}$

Step6: Calculate the result for c

Since $\cos(0)=1$ and $\sin(0)=0$, we have $\frac{0 + 1}{0+1}=1$

Step7: Analyze the limit as x approaches infinity for d

For $\lim_{x
ightarrow\infty}(\frac{-3}{x^{4}}+1)$, as $x
ightarrow\infty$, $\frac{-3}{x^{4}}
ightarrow0$. So the limit is $0 + 1=1$

Step8: Analyze the right - hand limit for e

For $\lim_{x
ightarrow5^{+}}\frac{-1}{(x - 5)^{2}}$, as $x
ightarrow5^{+}$, $(x - 5)^{2}
ightarrow0^{+}$, so $\frac{-1}{(x - 5)^{2}}
ightarrow-\infty$

Step9: Analyze the limit for f

The expression $\lim\frac{-1}{-5}$ is a constant function, so the limit is $\frac{1}{5}$

Step10: Analyze the right - hand limit for g

For $\lim_{x
ightarrow5^{+}}\frac{1}{x - 5}$, as $x
ightarrow5^{+}$, $x-5
ightarrow0^{+}$, so $\frac{1}{x - 5}
ightarrow+\infty$

Step11: Analyze the left - hand limit for h

For $\lim_{x
ightarrow5^{-}}\frac{1}{x - 5}$, as $x
ightarrow5^{-}$, $x - 5
ightarrow0^{-}$, so $\frac{1}{x - 5}
ightarrow-\infty$

Step12: Substitute the value for i

For $\lim_{\theta
ightarrow0}\frac{\cos\theta}{1}$, substitute $\theta = 0$, we get $\cos(0)=1$

Answer:

a. $5$
b. $4$
c. $1$
d. $1$
e. $-\infty$
f. $\frac{1}{5}$
g. $+\infty$
h. $-\infty$
i. $1$