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Evaluate each limit, if it exists, using any appropriate technique. 1.) The limit as u...

Question:

Evaluate each limit, if it exists, using any appropriate technique.

1.) {eq}\lim_{u \to 4}(u^2-\frac{16}{u^3}-64) {/eq}

2.) {eq}\lim_{x \to 1} (x^3+ x^2-5x+\frac3{x^2}-2x+1) {/eq}

3.) {eq}\lim_{u \to 1}( \sqrt{1+u} - \sqrt{1+\frac{u^2}{u-1}} ) {/eq}

4.) {eq}\lim_{x \to 3^+}( x \sqrt{x-3}) {/eq}

Limits:

From the behavior of the expression, the function can diverge at any input value or converge to a function value when the limit of the expression of function is taken from any input value.

Answer and Explanation:

1. Find: {eq}\lim_{u \to 4} (u^2-\frac{16}{u^3}-64) {/eq}


The strategy here is to simplify the expression before taking the limit, if possible.


{eq}\begin{align*} \lim_{u \to 4} (u^2-\frac{16}{u^3}-64) &= (4)^2-\frac{16}{(4)^3}-64 \text{ [Expression already simplified]} \\ &= 16-\frac{16}{64}-64 \\ &= -\frac{16}{64}-48 \\ &= -\frac{1}{4}-\frac{192}{4} \\ &= -\frac{193}{4} \\ \end{align*} {/eq}


2. Find: {eq}\lim_{x \to 1} (x^3+x^2-5x+\frac{3}{x^2}-2x+1) {/eq}


The strategy here is to simplify the expression before taking the limit, if possible.


{eq}\begin{align*} \lim_{x \to 1} (x^3+x^2-5x+\frac{3}{x^2}-2x+1) &= (1)^3+(1)^2-5(1)+\frac{3}{(1)^2}-2(1)+1 \text{ [Expression already simplified]} \\ &= 1+1-5+\frac{3}{1}-2+1 \\ &= 3-4 \\ &= -1 \\ \end{align*} {/eq}


3. Find: {eq}\lim_{u \to 1} (\sqrt{1+u}-\sqrt{1+\frac{u^2}{u-1}}) {/eq}


Since the expression is undefined at {eq}u = 1 {/eq} (by inspection of the denominator in the second term), the strategy is take to take the two one-sided limits that path can take to reach {eq}u = 1 {/eq}. If these two one-sided limits converge to the same function value, the limit as {eq}u {/eq} approaches {eq}1 {/eq} exists at that same function value.


For {eq}\lim_{u \to 1^+} (\sqrt{1+u}-\sqrt{1+\frac{u^2}{u-1}}) {/eq}:


{eq}\begin{align*} \lim_{u \to 1^+} (\sqrt{1+u}-\sqrt{1+\frac{u^2}{u-1}}) &= \lim _{u\to \:1^+}\left(\sqrt{1+u}\right)-\lim _{u\to \:1^+}\left(\sqrt{1+\frac{u^2}{u-1}}\right) \\ &= \sqrt{1+1}-\sqrt{\lim _{u\to \:1^+}\left(1+\frac{u^2}{u-1}\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^+}\left(1+\lim _{u\to \:1^+}\left(\frac{u^2}{u-1}\right)\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^+}\left(1+\lim _{u\to \:1^+}\left(u^2\right)\cdot \lim _{u\to \:1^+}\left(\frac{1}{u-1}\right)\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^+}\left(1+\lim _{u\to \:1^+}\left(u^2\right)\cdot \lim _{u\to \:1^+}\left(\frac{1}{u-1}\right)\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^+}\left(1+(1)^2\cdot \infty\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^+}\left(1+1\cdot \infty\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^+}\left(1+\infty\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^+}\left(\infty\right)} \\ &= \sqrt{2}-\sqrt{\infty} \\ &= \sqrt{2}-\infty \\ &= -\infty \\ \end{align*} {/eq}


For {eq}\lim_{u \to 1^-} (\sqrt{1+u}-\sqrt{1+\frac{u^2}{u-1}}) {/eq}:


{eq}\begin{align*} \lim_{u \to 1^-} (\sqrt{1+u}-\sqrt{1+\frac{u^2}{u-1}}) &= \lim _{u\to \:1^-}\left(\sqrt{1+u}\right)-\lim _{u\to \:1^-}\left(\sqrt{1+\frac{u^2}{u-1}}\right) \\ &= \sqrt{1+1}-\sqrt{\lim _{u\to \:1^-}\left(1+\frac{u^2}{u-1}\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^-}\left(1+\lim _{u\to \:1^-}\left(\frac{u^2}{u-1}\right)\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^-}\left(1+\lim _{u\to \:1^-}\left(u^2\right)\cdot \lim _{u\to \:1^-}\left(\frac{1}{u-1}\right)\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^-}\left(1+\lim _{u\to \:1^-}\left(u^2\right)\cdot \lim _{u\to \:1^-}\left(\frac{1}{u-1}\right)\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^-}\left(1+(1)^2\cdot -\infty\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^-}\left(1+1\cdot -\infty\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^-}\left(1-\infty\right)} \\ &= \sqrt{2}-\sqrt{\lim _{u\to \:1^-}\left(-\infty\right)} \\ &= \sqrt{2}-\sqrt{-\infty} \\ \end{align*} {/eq}


Because the two one-sided limits don't converge to the same function value, {eq}\lim_{u \to 1^-} (\sqrt{1+u}-\sqrt{1+\frac{u^2}{u-1}}) {/eq} does not exist.


4. Find: {eq}\lim_{x \to 3^+} (x\sqrt{x-3}) {/eq}


The strategy here is to simplify the expression before taking the limit, if possible.


{eq}\begin{align*} \lim_{x \to 3^+} (x\sqrt{x-3}) &= (3)\sqrt{3-3} \text{ [Limit already simplified]} \\ &= (3)\sqrt{0} \\ &= (3)(0) \\ &= 0 \\ \end{align*} {/eq}


Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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