Computing Limits using Laws
Find the limit if it exists.
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(a) \(\displaystyle{\lim_{x \to -1} \,
\dfrac{x+1}{x-1}}\)
(b) \(\displaystyle{\lim_{x \to -1} \, \dfrac{2x+2}{x^2-1}}\) solution
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(a) \(\displaystyle{\lim_{x \to 0} \,
\dfrac{x^2-4x}{x^3+3x}}\)
(b) \(\displaystyle{\lim_{x \to 2} \frac{x^2-2x}{x^2-x-2} }\) solution
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(a) \(\displaystyle{\lim_{x \to -1} 5x^3-3x^2+7x-11}\)
(b) \(\displaystyle{\lim_{x \to 3} \frac{x^2-2x-3}{x-3} }\) solution
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(a) \(\displaystyle{\lim_{h\to 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}}\)
(b) \(\displaystyle{\lim_{h \to 0} \frac{\sqrt{9+h}-3}{h} }\) solution
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\(\displaystyle{\lim_{h \rightarrow 0} \:
\frac{\sqrt{25+h}-5}{h}} \) solution
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\(\displaystyle{ \lim_{x \rightarrow 4} \:
\frac{\sqrt{x}-2}{x-4}} \) solution
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\(\displaystyle{\lim_{h \rightarrow 0}\:
\frac{(2+h)^2-4}{h}} \) solution
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\(\displaystyle{\lim_{x \rightarrow 2}\:
\frac{\frac{1}{2}-\frac{1}{x}}{2-x}} \) solution1
solution2
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\(\displaystyle{\lim_{x \rightarrow 4}\:
\frac{\frac{1}{3}-\frac{1}{\sqrt{x+5}}}{x-4}} \) solution
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\(\displaystyle{ \lim_{x \rightarrow 16}\:
\frac{16x-x^2}{4-\sqrt{x}}} \) solution
\(\displaystyle{\lim_{h \rightarrow 0}\:
\frac{\frac{1}{(a+h)^2}-\frac{1}{a^2}}{h}} \) solution
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\(\displaystyle{\lim_{x \rightarrow -2}\:
\frac{|x+2|}{3x+6}} \) solution
\(\displaystyle{\lim_{x \rightarrow 0}\: x^2\sin
\left(\frac{1}{x}\right)} \) solution
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\(\displaystyle{\lim_{x \rightarrow 0}\: x^3\cos \left(\frac{5}{x}\right)}\) solution
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If \( 2x\le f(x)\le x^4-x^2+2 \) for all \(x\), find the
value of \(\displaystyle{\lim_{x \rightarrow 1}\:
f(x)}. \) solution
Evaluate the limits or state that they do not exist. solution
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(a) \( \displaystyle{\lim_{x \to 3^-} \frac{2}{(x-3)^3}}\) (b) \( \displaystyle{\lim_{x \to 3^+ } \frac{2}{(x-3)^3}} \) (c) \( \displaystyle{\lim_{x \to 3} \frac{2}{(x-3)^3}} \)