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Calculus 1 : Calculus

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Example Questions

Example Question #1251 : Calculus

Solve for f{}'(c) using the Mean Value Theorem, rounded to the nearest hundredth place when

$f(x)=2x^4 - \frac{x^2}{7}$ on the interval $\left [ 1,3 \right ]$

Possible Answers:

79.43

75.24

80.65

83.91

Correct answer:

79.43

Explanation:

Mean Value Theorem (MVT) = $f{}'(c)= \frac{f(b)-f(a)}{b-a}$ on $\left [ a,b \right ]$

$f{}'(c)= \frac{(2(3)^4-\frac{(3)^2)}{7})- (2(1)^4-\frac{(1)^2}{7}))}{3-1}$

f{}'(c)= 79.43

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Example Question #1252 : Calculus

Evaluate the limit:

$Lim _{x \rightarrow -1} \frac{x^{^{2}}+4x+3}{x+1}$

Possible Answers:

Does Not Exist

Correct answer:

Explanation:

Attempting to evaluate directly (plug in -1 for ) results in the indeterminate form:

$\frac{0}{0}$

Further analysis is required:

$Lim _{x \rightarrow -1} \frac{x^{^{2}}+4x+3}{x+1}\dot{} = Lim _{x \rightarrow -1} \frac{(x+3)(x+1)}{x+1}\dot{} =Lim _{x \rightarrow -1}x+3$

This final form can be evaluated directly:

$Lim _{x \rightarrow -1} x+3 = -1 + 3=2$

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Example Question #1253 : Calculus

Evaluate the limit:

$Lim _{x \rightarrow 2} \frac{x^2 +5x-14}{x-2}$

Possible Answers:

Does Not Exist

Correct answer:

Explanation:

Attempting to evaluate directly (plug in 2 for ) results in the indeterminate form:

$\frac{0}{0}$

Further analysis is required:

$Lim _{x \rightarrow 2} \frac{x^2 +5x-14}{x-2}\dot{} = Lim _{x \rightarrow 2} \frac{(x+7)(x-2)}{x-2}\dot{} =Lim _{x \rightarrow 2}x+7$

This final form can be evaluated directly:

$Lim _{x \rightarrow 2} x+7 = 2+7=9$

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Example Question #224 : Functions

Evaluate the limit:

$Lim _{x \rightarrow 3} \frac{x^2 + 7x +6}{x-3}$

Possible Answers:

Does Not Exist

Correct answer:

Does Not Exist

Explanation:

We evaluate the limit directly (plug in 3 for ) and obtain:

$\frac{36}{0}$

from which we determine that the the function has a vertical asymptote at this point (it goes off to positive or negative infinity). The limit Does Not Exist.

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Example Question #1254 : Calculus

Given:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

Evaluate the limit:

$Lim _{x \rightarrow 0} xsecx$

Possible Answers:

$\pi$

Does Not Exist

Correct answer:

Explanation:

This limit can be evaluated directly.

Recall that $secx = \frac{1}{cosx}$

So:

$Lim _{x \rightarrow 0} xsecx = Lim _{x \rightarrow 0} \frac{x}{cosx}= \frac{0}{1}=0$

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Example Question #1255 : Calculus

Given:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

Evaluate the limit:

$Lim _{x \rightarrow 0} xcsc(2\pi x)$

Possible Answers:

$2\pi$

$\frac{1}{2\pi}$

$\frac{1}{2}$

DoesNotExist

Correct answer:

$\frac{1}{2\pi}$

Explanation:

First observe that $xcsc(2\pi x) = \frac{x}{sin2\pi x}$

Multiplying by $\frac{2\pi x}{2\pi x}$ we obtain:

$Lim _{x \rightarrow 0} \frac{2 \pi x*x}{2\pi x*sin2\pi x} = Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*\frac{x}{2\pi x}$

Limit of product is the product of limits:

$Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*\frac{x}{2\pi x}=Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*Lim _{x \rightarrow 0}\frac{x}{2\pi x}$

From the Pre-Question Text:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

So:

$Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*Lim _{x \rightarrow 0}\frac{x}{2\pi x}=1*\frac{1}{2\pi}=\frac{1}{2\pi}$

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Example Question #1256 : Calculus

Given:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^n=e$

Evaluate the limit:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{2n}$

Possible Answers:

DoesNotExist

e^2

$\frac{e}{2}$

Correct answer:

e^2

Explanation:

Recalling properties of exponents:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{2n}=Lim _{n \rightarrow \infty}\left ( \left (1+\frac{1}{n} \right )^{n} \right )^2=Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{n}\left (1+\frac{1}{n} \right )^{n}$

Limit of product is the product of the limits:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{n}\left (1+\frac{1}{n} \right )^{n}=Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{n} Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{n}$

$= \left (Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{n} \right )^2$

And from the Pre-Question Text:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^n=e$

So:

$\left (Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^{n} \right )^2 = e^2$

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Example Question #1257 : Calculus

Given:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^n=e$

Evaluate the limit:

$Lim _{n \rightarrow 0} \left (1+n \right )^{\frac{1}{n}}$

Possible Answers:

DoesNotExist

$\frac{1}{e}$

Correct answer:

Explanation:

To solve this problem we use the variable substitution:

$m = \frac{1}{n}$

Obtaining:

$Lim _{n \rightarrow 0} \left (1+n \right )^{\frac{1}{n}}=Lim _{m \rightarrow \infty} \left (1+\frac{1}{m} \right )^{m}$

And from the Pre-Question Text:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^n=e$

So:

$Lim _{m \rightarrow \infty} \left (1+\frac{1}{m} \right )^{m}=e$

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Example Question #1258 : Calculus

Given:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^n=e$

Evaluate the limit:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{2n} \right )^{n}$

Possible Answers:

$\frac{e}{2}$

DoesNotExist

$\sqrt[]{e}$

Correct answer:

$\sqrt[]{e}$

Explanation:

To solve this problem we use the variable substitution:

m = 2n

Obtaining:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{2n} \right )^{n} = Lim _{m \rightarrow \infty} \left (1+\frac{1}{m} \right )^{\frac{m}{2}} = L$

We then observe that:

$L^2 =Lim _{m \rightarrow \infty} \left (1+\frac{1}{m} \right )^{\frac{m}{2}}*Lim _{m \rightarrow \infty} \left (1+\frac{1}{m} \right )^{\frac{m}{2}}$

Product of limits is the limit of the products:

$= Lim _{m \rightarrow \infty} \left (1+\frac{1}{m} \right )^\frac{m}{2}*\left (1+\frac{1}{m} \right )^\frac{m}{2}=Lim _{m \rightarrow \infty}\left (1+\frac{1}{m} \right )^m$

And from the Pre-Question Text:

$Lim _{n \rightarrow \infty} \left (1+\frac{1}{n} \right )^n=e$

So:

$L^2 = Lim _{m \rightarrow \infty}\left (1+\frac{1}{m} \right )^m=e$

$L^2 = e \rightarrow L=e^{\frac{1}{2}}=\sqrt[]{e}$

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Example Question #1259 : Calculus

Given:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

Evaluate the limit:

$Lim _{x \rightarrow 0} \frac{x-xcosx}{sin^{2}x}$

Possible Answers:

- $-\frac{\pi}{2}$

DoesNotExist

$\pi^2$

Correct answer:

Explanation:

Multiplying by $\frac{x}{x}$ we obtain:

$Lim _{x \rightarrow 0} \frac{x*(x-xcosx)}{x*sin^{2}x} = Lim _{x \rightarrow 0} \frac{x}{sinx}*\frac{x}{sinx}* \frac{1-cosx}{x}$

Limit of product is the product of limits:

$Lim _{x \rightarrow 0} \frac{x}{sinx}*\frac{x}{sinx}* \frac{1-cosx}{x} = Lim _{x \rightarrow 0} \frac{x}{sinx}*Lim _{x \rightarrow 0}\frac{x}{sinx}*Lim _{x \rightarrow 0} \frac{1-cosx}{x}$

And from the Pre-Question Text:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$ and $Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

So:

$Lim _{x \rightarrow 0} \frac{x}{sinx}*Lim _{x \rightarrow 0}\frac{x}{sinx}*Lim _{x \rightarrow 0} \frac{1-cosx}{x}=1*1*0=0$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1251 : Calculus

Example Question #1252 : Calculus

Example Question #1253 : Calculus

Example Question #224 : Functions

Example Question #1254 : Calculus

Example Question #1255 : Calculus

Example Question #1256 : Calculus

Example Question #1257 : Calculus

Example Question #1258 : Calculus

Example Question #1259 : Calculus

All Calculus 1 Resources

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