Limits with indeterminate forms exercises

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Here is an opportunity for you to practice evaluating limits with indeterminate forms.

$\lim _{z \to 0} \frac {z^2+z}{z^3-z} = \answer {-1}$

$\lim _{q \to 0} \cfrac {q}{\frac {1}{q} - \frac {1}{q-1}} = \answer {0}$

$\lim _{x \to 4} \frac {x-16}{\sqrt {x} + 4} = \answer {-2}$

$\lim _{c \to 1} \cfrac {\frac {1}{c-1}}{\frac {1}{2c-2}} = \answer {2}$

$\lim _{\alpha \to \pi } \sin (\alpha ) \cos (\alpha ) \cot (\alpha ) = \answer {1}$

If you are struggling with this problem, try to re-write $\cot (\alpha )$ in terms of $\sin (\alpha )$ and $\cos (\alpha )$ . If you are unsure how to do this, you may want to review the definition of cotangent here or here.

$\lim _{a \to 0} \cfrac {\sqrt {a+4} -2}{a} = \answer {\frac {1}{4}}$

$\lim _{x \to -5} \frac {x^2 + 4x -5}{x^2 + 7x + 10} = \answer {2}$

$\displaystyle \lim _{\theta \to 3\pi /2}\frac {\sin ^2(\theta )\cos (\theta ) + \cos ^3(\theta )}{\cos (\theta )} = \answer {1}$

$\lim _{l \to -3} \cfrac {1+\frac {3}{l}}{l+3} = \answer {\frac {-1}{3}}$

Consider the statement below, and then indicate whether it is sometimes, always, or never true.

‘‘If $\displaystyle \lim _{x \to -5} f(x)$ is of the form $\frac {0}{0}$ , then $\displaystyle \lim _{x\to -5} f(x)$ does not exist.”

This statement is

sometimesalwaysnever true.