Calculus Examples

$lim_{t \to \infty} \frac{e^{3 t} + t^{2}}{2 e^{3 t} - t}$

Step 1

Divide the numerator and denominator by the fastest growing term in the denominator.

$lim_{t \to \infty} \frac{\frac{e^{3 t}}{e^{3 t}} + \frac{t^{2}}{e^{3 t}}}{\frac{2 e^{3 t}}{e^{3 t}} - \frac{t}{e^{3 t}}}$

Step 2

Evaluate the limit.

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Step 2.1

Cancel the common factor of $e^{3 t}$ .

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Step 2.1.1

Cancel the common factor.

$lim_{t \to \infty} \frac{\frac{e^{3 t}}{e^{3 t}} + \frac{t^{2}}{e^{3 t}}}{\frac{2 e^{3 t}}{e^{3 t}} - \frac{t}{e^{3 t}}}$

Step 2.1.2

Rewrite the expression.

$lim_{t \to \infty} \frac{1 + \frac{t^{2}}{e^{3 t}}}{\frac{2 e^{3 t}}{e^{3 t}} - \frac{t}{e^{3 t}}}$

Step 2.2

Cancel the common factor of $e^{3 t}$ .

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Step 2.2.1

Cancel the common factor.

$lim_{t \to \infty} \frac{1 + \frac{t^{2}}{e^{3 t}}}{\frac{2 e^{3 t}}{e^{3 t}} - \frac{t}{e^{3 t}}}$

Step 2.2.2

Divide $2$ by $1$ .

$lim_{t \to \infty} \frac{1 + \frac{t^{2}}{e^{3 t}}}{2 - \frac{t}{e^{3 t}}}$

Step 2.3

Split the limit using the Limits Quotient Rule on the limit as $t$ approaches $\infty$ .

$\frac{lim_{t \to \infty} 1 + \frac{t^{2}}{e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 2.4

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $\infty$ .

$\frac{lim_{t \to \infty} 1 + lim_{t \to \infty} \frac{t^{2}}{e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 2.5

Evaluate the limit of $1$ which is constant as $t$ approaches $\infty$ .

$\frac{1 + lim_{t \to \infty} \frac{t^{2}}{e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 3.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{1 + \frac{lim_{t \to \infty} t^{2}}{lim_{t \to \infty} e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{1 + \frac{\infty}{lim_{t \to \infty} e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.1.3

Since the exponent $3 t$ approaches $\infty$ , the quantity $e^{3 t}$ approaches $\infty$ .

$\frac{1 + \frac{\infty}{\infty}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{1 + \frac{\infty}{\infty}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.2

Since $\frac{\infty}{\infty}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{t \to \infty} \frac{t^{2}}{e^{3 t}} = lim_{t \to \infty} \frac{\frac{d}{d t} [t^{2}]}{\frac{d}{d t} [e^{3 t}]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$\frac{1 + lim_{t \to \infty} \frac{\frac{d}{d t} [t^{2}]}{\frac{d}{d t} [e^{3 t}]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{\frac{d}{d t} [e^{3 t}]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 3 t$ .

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Step 3.3.3.1

To apply the Chain Rule, set $u$ as $3 t$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{\frac{d}{d u} [e^{u}] \frac{d}{d t} [3 t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{e^{u} \frac{d}{d t} [3 t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.3.3

Replace all occurrences of $u$ with $3 t$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{e^{3 t} \frac{d}{d t} [3 t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.4

Since $3$ is constant with respect to $t$ , the derivative of $3 t$ with respect to $t$ is $3 \frac{d}{d t} [t]$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{e^{3 t} \cdot 3 \frac{d}{d t} [t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.5

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{e^{3 t} \cdot 3 \cdot 1}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.6

Multiply $3$ by $1$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{e^{3 t} \cdot 3}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 3.3.7

Move $3$ to the left of $e^{3 t}$ .

$\frac{1 + lim_{t \to \infty} \frac{2 t}{3 e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 4

Move the term $\frac{2}{3}$ outside of the limit because it is constant with respect to $t$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{t}{e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5

Apply L'Hospital's rule.

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Step 5.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 5.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{1 + \frac{2}{3} \cdot \frac{lim_{t \to \infty} t}{lim_{t \to \infty} e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{1 + \frac{2}{3} \cdot \frac{\infty}{lim_{t \to \infty} e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.1.3

Since the exponent $3 t$ approaches $\infty$ , the quantity $e^{3 t}$ approaches $\infty$ .

$\frac{1 + \frac{2}{3} \cdot \frac{\infty}{\infty}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{1 + \frac{2}{3} \cdot \frac{\infty}{\infty}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.2

$lim_{t \to \infty} \frac{t}{e^{3 t}} = lim_{t \to \infty} \frac{\frac{d}{d t} [t]}{\frac{d}{d t} [e^{3 t}]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{\frac{d}{d t} [t]}{\frac{d}{d t} [e^{3 t}]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{\frac{d}{d t} [e^{3 t}]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 3 t$ .

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Step 5.3.3.1

To apply the Chain Rule, set $u$ as $3 t$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{\frac{d}{d u} [e^{u}] \frac{d}{d t} [3 t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{e^{u} \frac{d}{d t} [3 t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.3.3

Replace all occurrences of $u$ with $3 t$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{e^{3 t} \frac{d}{d t} [3 t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.4

Since $3$ is constant with respect to $t$ , the derivative of $3 t$ with respect to $t$ is $3 \frac{d}{d t} [t]$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{e^{3 t} \cdot 3 \frac{d}{d t} [t]}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.5

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{e^{3 t} \cdot 3 \cdot 1}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.6

Multiply $3$ by $1$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{e^{3 t} \cdot 3}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 5.3.7

Move $3$ to the left of $e^{3 t}$ .

$\frac{1 + \frac{2}{3} lim_{t \to \infty} \frac{1}{3 e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 6

Move the term $\frac{1}{3}$ outside of the limit because it is constant with respect to $t$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} lim_{t \to \infty} \frac{1}{e^{3 t}}}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 7

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{3 t}}$ approaches $0$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{lim_{t \to \infty} 2 - \frac{t}{e^{3 t}}}$

Step 8

Evaluate the limit.

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Step 8.1

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $\infty$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{lim_{t \to \infty} 2 - lim_{t \to \infty} \frac{t}{e^{3 t}}}$

Step 8.2

Evaluate the limit of $2$ which is constant as $t$ approaches $\infty$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{t}{e^{3 t}}}$

Step 9

Apply L'Hospital's rule.

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Step 9.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 9.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - \frac{lim_{t \to \infty} t}{lim_{t \to \infty} e^{3 t}}}$

Step 9.1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - \frac{\infty}{lim_{t \to \infty} e^{3 t}}}$

Step 9.1.3

Since the exponent $3 t$ approaches $\infty$ , the quantity $e^{3 t}$ approaches $\infty$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - \frac{\infty}{\infty}}$

Step 9.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - \frac{\infty}{\infty}}$

Step 9.2

$lim_{t \to \infty} \frac{t}{e^{3 t}} = lim_{t \to \infty} \frac{\frac{d}{d t} [t]}{\frac{d}{d t} [e^{3 t}]}$

Step 9.3

Find the derivative of the numerator and denominator.

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Step 9.3.1

Differentiate the numerator and denominator.

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{\frac{d}{d t} [t]}{\frac{d}{d t} [e^{3 t}]}}$

Step 9.3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{\frac{d}{d t} [e^{3 t}]}}$

Step 9.3.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 3 t$ .

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Step 9.3.3.1

To apply the Chain Rule, set $u$ as $3 t$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{\frac{d}{d u} [e^{u}] \frac{d}{d t} [3 t]}}$

Step 9.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{e^{u} \frac{d}{d t} [3 t]}}$

Step 9.3.3.3

Replace all occurrences of $u$ with $3 t$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{e^{3 t} \frac{d}{d t} [3 t]}}$

Step 9.3.4

Since $3$ is constant with respect to $t$ , the derivative of $3 t$ with respect to $t$ is $3 \frac{d}{d t} [t]$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{e^{3 t} \cdot 3 \frac{d}{d t} [t]}}$

Step 9.3.5

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{e^{3 t} \cdot 3 \cdot 1}}$

Step 9.3.6

Multiply $3$ by $1$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{e^{3 t} \cdot 3}}$

Step 9.3.7

Move $3$ to the left of $e^{3 t}$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - lim_{t \to \infty} \frac{1}{3 e^{3 t}}}$

Step 10

Move the term $\frac{1}{3}$ outside of the limit because it is constant with respect to $t$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - \frac{1}{3} lim_{t \to \infty} \frac{1}{e^{3 t}}}$

Step 11

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{3 t}}$ approaches $0$ .

$\frac{1 + \frac{2}{3} \cdot \frac{1}{3} \cdot 0}{2 - \frac{1}{3} \cdot 0}$

Step 12

Simplify the answer.

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Step 12.1

Simplify the numerator.

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Step 12.1.1

Multiply $\frac{2}{3} \cdot \frac{1}{3}$ .

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Step 12.1.1.1

Multiply $\frac{2}{3}$ by $\frac{1}{3}$ .

$\frac{1 + \frac{2}{3 \cdot 3} \cdot 0}{2 - \frac{1}{3} \cdot 0}$

Step 12.1.1.2

Multiply $3$ by $3$ .

$\frac{1 + \frac{2}{9} \cdot 0}{2 - \frac{1}{3} \cdot 0}$

Step 12.1.2

Multiply $\frac{2}{9}$ by $0$ .

$\frac{1 + 0}{2 - \frac{1}{3} \cdot 0}$

Step 12.1.3

Add $1$ and $0$ .

$\frac{1}{2 - \frac{1}{3} \cdot 0}$

Step 12.2

Simplify the denominator.

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Step 12.2.1

Multiply $- \frac{1}{3} \cdot 0$ .

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Step 12.2.1.1

Multiply $0$ by $- 1$ .

$\frac{1}{2 + 0 (\frac{1}{3})}$

Step 12.2.1.2

Multiply $0$ by $\frac{1}{3}$ .

$\frac{1}{2 + 0}$

Step 12.2.2

Add $2$ and $0$ .

$\frac{1}{2}$