Section 1.2 Limits
Evaluate the following limits. Use limit theorems, not ε - δ techniques. If any of them fail to exist, say so and say why.lim
\displaystyle \displaystyle \lim _{x\to 10} \frac{x^2-99}{x-10}
\displaystyle \displaystyle \lim _{x\to 10} \frac{x^2-100}{x-9}
\ds \lim _{x\to 10} f(x)\text{,} where f(x)=x^2 for all x\not= 10\text{,} but f(10)=99\text{.}
\displaystyle \displaystyle \lim _{x\to 10}\sqrt{-x^2+20x-100}
\displaystyle \displaystyle \lim _{x\to -4} \frac{x^2-16}{x+4}\ln |x|
\displaystyle \displaystyle \lim _{x\to \infty} \frac{x^2}{e^{4x}-1-4x}
\displaystyle \displaystyle \lim _{x\to -\infty} \frac{3x^6-7x^5+x}{5x^6+4x^5-3}
\displaystyle \displaystyle \lim _{x\to -\infty} \frac{5x^7-7x^5+1}{2x^7+6x^6-3}
\displaystyle \displaystyle \lim _{x\to -\infty} \frac{2x+3x^3}{x^3+2x-1}
\displaystyle \displaystyle \lim _{x\to -\infty} \frac{5x+2x^3}{x^3+x-7}
\displaystyle \lim _{x\to \infty} \frac{ax^{17}+bx}{cx^{17}-dx^3}\text{,} a,b,c,d\not=0
\displaystyle \displaystyle \lim _{x\to \infty} \frac{3x+|1-3x|}{1-5x}
\displaystyle \displaystyle \lim _{x\to -\infty} \frac{\sqrt{x^6-3}}{\sqrt{x^6+5}}
\displaystyle \displaystyle \lim _{u\to \infty} \frac{u}{\sqrt{u^2+1}}
\displaystyle \displaystyle \lim _{x\to \infty} \frac{1+3x}{\sqrt{2x^2+x}}
\displaystyle \displaystyle \lim _{x\to \infty} \frac{\sqrt{4x^2+3x}-7}{7-3x}
\displaystyle \displaystyle \lim _{x\to -\infty} \frac{\sqrt{x^2-9}}{2x-1}
\displaystyle \displaystyle \lim _{x\to 1^+} \frac{\sqrt{x-1}}{x^2-1}
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Let \ds f(x)=\left\{ \begin{array}{lll} \frac{x^2-1}{|x-1|}\amp \mbox{if } \amp x\not= 1,\\ 4\amp \mbox{if } \amp x= 1. \end{array} \right.
Find \displaystyle \lim _{x\to 1^-}f(x)\text{.}
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Let F(x)=\frac{2x^2-3x}{|2x-3|}\text{.}
Find \displaystyle \lim _{x\to 1.5^+}F(x)\text{.}
Find \displaystyle \lim _{x\to 1.5^-}F(x)\text{.}
Does \displaystyle \lim _{x\to 1.5}F(x) exist? Provide a reason.
\displaystyle \displaystyle \lim _{x\to -2} \frac{2-|x|}{2+x}
\displaystyle \displaystyle \lim _{x\to 2^-} \frac{|x^2-4|}{10-5x}
\displaystyle \displaystyle \lim _{x\to 4^-} \frac{|x-4|}{(x-4)^2}
\displaystyle \displaystyle \lim _{x\to 8} \frac{(x-8)(x+2)}{|x-8|}
\displaystyle \displaystyle \lim _{x\to 2} \left(\frac{1}{x^2+5x+6}-\frac{1}{x-2}\right)
\displaystyle \displaystyle \lim _{x\to -1} \frac{x^2-x-2}{3x^2-x-1}
\displaystyle \displaystyle \lim _{x\to 16}\frac{\sqrt{x}-4}{x-16}
\displaystyle \displaystyle \lim _{x\to 8}\frac{\sqrt[3]{x}-2}{x-8}
\displaystyle \displaystyle \lim _{x\to 4} \frac{2-\sqrt{x}}{4x-x^2}
\displaystyle \displaystyle \lim _{x\to 0} \frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}
Find constants a and b such that \displaystyle \lim _{x\to 0}\frac{\sqrt{ax+b}-2}{x}=1\text{.}
\displaystyle \displaystyle \lim _{x\to 5}e^{ \frac{x-5}{\sqrt{x-1}-2}}
\displaystyle \displaystyle \lim _{x\to 7}e^{ \frac{\sqrt{x+2}-3}{x-7}}
\displaystyle \displaystyle \lim _{t\to 0} \frac{\sqrt{\sin t +1}-1}{t}
\displaystyle \displaystyle \lim _{x\to 8}\frac{x^{1/3}-2}{x-8}
\displaystyle \displaystyle \lim _{x\to \infty }\left( \sqrt{x^2+x}-x\right)
\displaystyle \displaystyle \lim _{x\to -\infty }\left( \sqrt{x^2+5x}-\sqrt{x^2+2x}\right)
\displaystyle \displaystyle \lim _{x\to \infty }\left( \sqrt{x^2-x+1}-\sqrt{x^2+1}\right)
\displaystyle \displaystyle \lim _{x\to \infty }\left( \sqrt{x^2+3x-2}-x\right)
Is there a number b such that \ds \lim _{x\to -2}\frac{bx^2+15x+15+b}{x^2+x-2} exists? If so, find the value of b and the value of the limit.
Determine the value of a so that \ds f(x)=\frac{x^2+ax+5}{x+1} has a slant asymptote y=x+3\text{.}
Prove that f(x)=\frac{\ln x}{x} has a horizontal asymptote y=0\text{.}
Let I be an open interval such that 4\in I and let a function f be defined on a set D=I\backslash \{ 4\}\text{.} Evaluate \displaystyle \lim _{x\to 4}f(x)\text{,} where x+2\leq f(x)\leq x^2-10 for all x\in D\text{.}
Evaluate \displaystyle \lim _{x\to 1}f(x)\text{,} where 2x-1\leq f(x)\leq x^2 for all x in the interval (0,2)\text{.}
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Use the squeeze theorem to show that
\displaystyle \lim _{x\to 0}x^4\sin\left( \frac{1}{x}\right) =0\text{,}
\displaystyle \lim _{x\to 0^+}\left( \sqrt{x}e^{\sin (1/x)}\right) =0\text{.}
Evaluate the following limits. If any of them fail to exist, say so and say why.
\displaystyle \displaystyle \lim _{x\to 0^+}\left[ (x^2+x)^{1/3}\sin \left( \frac{1}{x^2}\right)\right]
\displaystyle \displaystyle \lim _{x\to 0}x\sin \left( \frac{e}{x}\right)
\displaystyle \displaystyle \lim _{x\to 0}x\sin \left( \frac{1}{x^2}\right)
\displaystyle \displaystyle \lim _{x\to 0} \sqrt{x^2+x}\cdot \sin\left(\frac{\pi}{x}\right)
\displaystyle \displaystyle \lim _{x\to 0} x\cos^2\left(\frac{1}{x^2}\right)
\displaystyle \displaystyle \lim _{x\to \pi /2^+}\frac{x}{\cot x}
\displaystyle \displaystyle \lim _{x\to 0}\frac{1-e^{-x}}{1-x}
\displaystyle \displaystyle \lim _{x\to 0} \frac{e^{2x}-1-2x}{x^2}
\displaystyle \displaystyle \lim _{x\to 2} \frac{e^x-e^2}{\cos\left(\frac{\pi x}{2}\right)+1}
\displaystyle \displaystyle \lim _{x\to 1} \frac{x^2-1}{e^{1-x^7}-1}
\displaystyle \displaystyle \lim _{x\to 0}\frac{e^{-x^2}\cos (x^2)}{x^2}
\displaystyle \displaystyle \lim _{x\to 1}\frac{x^{76}-1}{x^{45}-1}
\displaystyle \lim _{x\to 1} \frac{x^a-1}{x^b-1}\text{,} a,b\not=0
\displaystyle \displaystyle \lim _{x\to 0}\frac{(\sin x)^{100}}{x^{99}\sin 2x}
\displaystyle \displaystyle \lim _{x\to 0}\frac{x^{100}\sin 7x}{(\sin x)^{99}}
\displaystyle \displaystyle \lim _{x\to 0}\frac{x^{100}\sin 7x}{(\sin x)^{101}}
\displaystyle \displaystyle \lim _{x\to 0}\frac{\arcsin 3x}{\arcsin 5x}
\displaystyle \displaystyle \lim _{x\to 0}\frac{\sin 3x}{\sin 5x}
\displaystyle \displaystyle \lim _{x\to 0} \frac{x^3\sin \left( \frac{1}{x^2}\right)}{\sin x}
\displaystyle \displaystyle \lim _{x\to 0}\frac{\sin x}{\sqrt{x\sin 4x}}
\displaystyle \displaystyle \lim _{x \to 0}\frac{1-\cos x}{x\sin x}
\displaystyle \displaystyle \lim _{\theta\to \frac{3\pi}{2}} \frac{\cos \theta +1}{\sin\theta}
\displaystyle \displaystyle \lim _{x\to \frac{\pi}{2}} \left(x-\frac{\pi}{2}\right)\tan x
\displaystyle \displaystyle \lim _{x\to \infty }x\tan (1/x)
\displaystyle \displaystyle \lim _{x\to 0}\left( \frac{1}{\sin x}-\frac{1}{x}\right)
\displaystyle \displaystyle \lim _{x\to 0}\frac{x- \sin x}{x^3}
\displaystyle \displaystyle \lim _{x\to 0} (\csc x-\cot x)
\displaystyle \displaystyle \lim _{x\to 0^+}(\sin x)(\ln \sin x)
\displaystyle \displaystyle \lim _{x\to \infty} \left(x\cdot \ln\frac{x-1}{x+1}\right)
\displaystyle \displaystyle \lim _{x\to \infty} \frac{e^{\frac{x}{10}}}{x^3}
\displaystyle \displaystyle \lim _{x\to \infty }\frac{\ln x}{\sqrt{x}}
\displaystyle \displaystyle \lim _{x\to \infty }\frac{\ln 3x}{x^2}
\displaystyle \displaystyle \lim _{x\to \infty }\frac{(\ln x)^2}{x}
\displaystyle \displaystyle \lim _{x\to 1 }\frac{\ln x}{x}
\displaystyle \displaystyle \lim _{x\to 0 }\frac{\ln (2+2x)-\ln 2}{x}
\displaystyle \displaystyle \lim _{x\to \infty }\frac{\ln ((2x)^{1/2})}{\ln ((3x)^{1/3})}
\displaystyle \displaystyle \lim _{x\to 0 }\frac{\ln (1+3x)}{2x}
\displaystyle \displaystyle \lim _{x\to 1 }\frac{\ln (1+3x)}{2x}
\displaystyle \displaystyle \lim _{\theta \to \frac{\pi }{2} ^+}\frac{\ln (\sin \theta)}{\cos \theta }
\displaystyle \displaystyle \lim _{x\to 1 }\frac{1-x+\ln x}{1+\cos (\pi x)}
\displaystyle \displaystyle \lim _{x\to 0 }\left( \frac{1}{x^2}-\frac{1}{\tan x}\right)
\displaystyle \displaystyle \lim _{x\to 0^+} \left(\frac{1}{x}-\frac{1}{e^x-1}\right)
\displaystyle \displaystyle\lim _{x\to 0}(\cosh x)^{\frac{1}{x^2}}
\displaystyle \displaystyle \lim _{x\to 0^+}(\cos x)^{\frac{1}{x}}
\displaystyle \displaystyle \lim _{x\to 0^+}(\cos x)^{\frac{1}{x^2}}
\displaystyle \displaystyle \lim _{x\to 0^+}x^{x}
\displaystyle \displaystyle \lim _{x\to 0^+}x^{\sqrt{x}}
\displaystyle \displaystyle \lim _{x\to 0^+} x^{\tan x}
\displaystyle \displaystyle \lim _{x\to 0^+}(\sin x)^{\tan x}
\displaystyle \displaystyle \lim _{x\to 0}(1+\sin x)^{\frac{1}{x}}
\displaystyle \displaystyle \lim _{x\to \infty }(x+\sin x)^{\frac{1}{x}}
\displaystyle \displaystyle \lim _{x\to \infty }x^{\frac{1}{x}}
\displaystyle \displaystyle \lim _{x\to \infty }\left( 1+ \frac{1}{x}\right) ^{2x}
\displaystyle \displaystyle \lim _{x\to \infty }\left( 1+\sin \frac{3}{x}\right) ^x
\displaystyle \displaystyle \lim _{x\to 0^+}(x+\sin x)^{\frac{1}{x}}
\displaystyle \displaystyle \lim _{x\to 0^+}\left( \frac{x}{x+1}\right) ^{x}
\displaystyle \displaystyle \lim _{x\to e^+}(\ln x)^{\frac{1}{x-e}}
\displaystyle \displaystyle \lim _{x\to e^+}(\ln x)^{\frac{1}{x}}
\displaystyle \displaystyle \lim _{x\to 0}e^{x\sin (1/x)}
\displaystyle \displaystyle \lim _{x\to 0}(1-2x)^{1/x}
\displaystyle \displaystyle \lim _{x\to 0^+}(1+7x)^{1/5x}
\displaystyle \displaystyle \lim _{x\to 0^+}(1+3x)^{1/8x}
\displaystyle \displaystyle \lim _{x\to 0}\left( 1+\frac{x}{2}\right) ^{3/x}
Let x_1=100\text{,} and for n\geq 1\text{,} let \displaystyle x_{n+1}=\frac{1}{2}\left(x_n+\frac{100}{x_n}\right)\text{.} Assume that \displaystyle L=\lim _{n\to \infty }x_n exists. Calculate L\text{.}
Find \displaystyle \lim _{x \to 0}\frac{1-\cos x}{x^2}\text{,} or show that it does not exist.
Find \displaystyle\lim _{x \to 2\pi }\frac{1-\cos x}{x^2}\text{,} or show that it does not exist.
Find \displaystyle \lim _{x \to -1}\arcsin x\text{,} or show that it does not exist.
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Compute the following limits or state why they do not exist:
\displaystyle \displaystyle \lim _{h\to 0}\frac{\sqrt[4]{16+h}-2}{2h}
\displaystyle \displaystyle \lim _{x\to 1}\frac{\ln x}{\sin (\pi x)}
\displaystyle \displaystyle \lim _{u\to \infty }\frac{u}{\sqrt{u^2+1}}
\displaystyle \displaystyle \lim _{x\to 0 }(1-2x)^{1/x}
\displaystyle \displaystyle \lim _{x\to 0 }\frac{(\sin x)^{100}}{x^{99}\sin (2x)}
\displaystyle \displaystyle \lim _{x\to \infty }\frac{1.01^x}{x^{100}}
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Find the following limits. If a limit does not exist, write 'DNE'. No justification necessary.
\displaystyle \displaystyle \lim _{x\to 0}\frac{(2+x)^{2016}-2^{2016}}{x}
\displaystyle \displaystyle \lim _{x\to \infty }(\sqrt{x^2+x}-x)
\displaystyle \displaystyle \lim _{x\to 0} \cot (3x)\sin (7x)
\displaystyle \displaystyle \lim _{x\to 0^+}x^x
\displaystyle \displaystyle \lim _{x\to \infty} \frac{x^2}{e^x}
\displaystyle \displaystyle \lim _{x\to 3}\frac{\sin x-x}{x^3}
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Evaluate the following limits, if they exist.
\displaystyle \lim _{x\to 0}\frac{f(x)}{|x|} given that \displaystyle \lim _{x\to 0}xf(x)=3\text{.}
\displaystyle \displaystyle \lim _{x\to 1} \frac{\sin (x-1)}{x^2+x-2}
\displaystyle \displaystyle \lim _{x\to -\infty }\frac{\sqrt{x^2+4x}}{4x+1}
\displaystyle \displaystyle \lim _{x\to \infty }\frac{\sqrt{x^4+2}}{x^4-4}
\displaystyle \displaystyle \lim _{x\to \infty} (e^x+x)^{1/x}
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Evaluate the following limits, if they exist.
\displaystyle \displaystyle \lim _{x\to 4}\left[ \frac{1}{\sqrt{x}-2}-\frac{4}{x-4}\right]
\displaystyle \displaystyle \lim _{x\to 1} \frac{x^2-1}{e^{1-x^2}-1}
\displaystyle \displaystyle \lim _{x\to 0}(\sin x)(\ln x)
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Evaluate the following limits. Use “\infty” or “-\infty” where appropriate.
\displaystyle \displaystyle \lim _{x\to 1^-}\frac{x+1}{x^2-1}
\displaystyle \displaystyle \lim _{x\to 0} \frac{\sin 6x}{2x}
\displaystyle \displaystyle \lim _{x\to 0}\frac{\sinh 2x}{xe^x}
\displaystyle \displaystyle \lim _{x\to 0^+}(x^{0.01}\ln x)
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Use the \varepsilon -- \delta definition of limits to prove that
\begin{equation*} \lim _{x\to 0}x^3=0\text{.} \end{equation*} Sketch an approximate graph of f(x)=2x^2 on [0,2]\text{.} Next, draw the points P(1,0) and Q(0,2)\text{.} When using the precise definition of \lim _{x\to 1}f(x)=2\text{,} a number \delta and another number \varepsilon are used. Show points on the graph which these values determine. (Recall that the interval determined by \delta must not be greater than a particular interval determined by \varepsilon\text{.})
Use the graph to find a positive number \delta so that whenever |x-1|\lt \delta it is always true that |2x^2-2|\lt \frac{1}{4}\text{.}
State exactly what has to be proved to establish this limit property of the function f\text{.}
Give an example of a function F=f+g such that the limits of f and g at a do not exist and that the limit of F at a exists.
If \ds \lim_{x\to a}[f(x)+g(x)]=2 and \ds \lim_{x\to a}[f(x)-g(x)]=1 find \ds \lim_{x\to a}[f(x)\cdot g(x)]\text{.}
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If f' is continuous, use L'Hospital's rule to show that
\begin{equation*} \displaystyle \lim _{h\to 0}\frac{f(x+h)-f(x-h)}{2h}=f'(x)\text{.} \end{equation*}Explain the meaning of this equation with the aid of a diagram.