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Evaluating a Limit Let $f(x)=x+x \sin x$ and $g(x)=x^{2}-4 .$(a) Show that $\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=0$(b) Show that $\lim _{x \rightarrow \infty} f(x)=\infty$ and $\lim _{x \rightarrow \infty} g(x)=\infty$(c) Evaluate the limit$$\lim _{x \rightarrow \infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}$$What do you notice?(d) Do your answers to parts (a) through (c) contradict L'Hopital's Rule? Explain your reasoning.

a) $$\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=0$$b) $$\lim _{x \rightarrow \infty} f(x)=\infty, \lim _{x \rightarrow \infty} g(x)=\infty$$c) $$\lim _{x \rightarrow \infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\infty$$

Calculus 1 / AB

Chapter 5

Logarithmic, Exponential, and Other Transcendental Functions

Section 6

Indeterminate Forms and L'Hopita's Rule

Functions

Differentiation

Integrals

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so part A. We have the limit as X approaches infinity of f of X plus x signed of X, divided by X squared minus four. So we divide the numerator and denominator expression by X square. We have thistle is equal to one over ax plus the sign of X divided by X all divided by one minus four over X squared. So this is equal to zero plus zero divided by one month Syria, which is equal to zero. Next for part B, we have the limit as X approaches infinity limit as X approaches infinity of f of X after lex, which is, um X plus X sign of X. And this is equal to infinity, plus infinity. So that's just infinity. And if we have the limit as X approaches infinity of G of axe, which is equal to X squared minus four, which is equal to infinity minus four, which is just equal to infinity. Finally, we have part C, which is if we take the limit as X approaches infinity off F prime divided by G crime. We're gonna get one plus x co sign of X plus the sign off acts old abided by two acts dividing the numerator and denominator by acts and then playing infinity. We're gonna get that This is equal to zero plus plus the co sign of infinity plus zero all divided by two. So since cold cider backs oscillating between negative one and one, this limit does not exist. Finally, for part D. So for party, if we can just All right, So for party, we're gonna say that Yes, because by definition of low petals rule, if the limit of F prime divided by G prime exists, then the limit off f divided by G off the limit of F kind. But G time for indeterminant forms such as this case is infinity divided by infinity.

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