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I'm trying to find $\lim_{x \rightarrow 0}f(x)$ and $\lim_{x \rightarrow 0}f(x)$ and $\lim_{x \rightarrow 0}\frac{f(x)}{x}$ using $\lim_{x \rightarrow 0}\frac{f(x)}{x^2}=1$. To find $\lim_{x \rightarrow 0}f(x)$ what I did was $$\lim_{x \rightarrow 0}f(x)=\lim_{x \rightarrow 0}\frac{f(x)}{x^2}\cdot\lim_{x \rightarrow 0}x^2=1\cdot \lim_{x \rightarrow 0}x^2=1\cdot 0=0$$ However I'm not so sure of finding $\lim_{x \rightarrow 0}\frac{f(x)}{x}$. What I tried doing was saying $$\lim_{x \rightarrow 0}\frac{f(x)}{x}=\lim_{x \rightarrow 0}\frac{f(x)}{x^2}\cdot \lim_{x \rightarrow 0}x=1\cdot 0=0$$ My solutions for both limits seems right but I'm just not sure how justified I am by multiplying the limits in both parts, respectively, by $\frac{\lim_{x \rightarrow 0}x^2}{\lim_{x \rightarrow 0}x^2}$ and $\frac{\lim_{x \rightarrow 0}x}{\lim_{x \rightarrow 0}x}$ even though they let me simplify the limits into the limits of functions whose limits I know.

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    $\begingroup$ You are only using: If $lim_{x \to 0}f(x)$ is limited and $\lim_{x\to 0}g(x)=0$ , then $\lim_{x \to 0}f(x)g(x)=0.$ And it's correct. $\endgroup$ – 7697 Oct 26 '17 at 18:55
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    $\begingroup$ It is certainly the case that if $\lim_{x\to c}f(x)$ exists and $\to A$, and if $\lim_{x\to c} g(x)$ exists and $\to B$, then $\lim_{x\to c} f(x)g(x)$ exists and $\to AB$. (note that $c$ here can also be $\infty$) You're just applying this principle with $B=0$. $\endgroup$ – Steven Stadnicki Oct 26 '17 at 18:57
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    $\begingroup$ You should add one more step like $$\lim_{x\to 0}f(x)=\lim_{x\to 0}x^{2}\cdot\frac{f(x)}{x^{2}}=\lim_{x\to 0}x^{2}\cdot\lim_{x\to 0}\frac{f(x)}{x^{2}}=\dots$$ The first equality is just plain algebra and second equality is better known as product rule of limits (part of algebra of limits). $\endgroup$ – Paramanand Singh Oct 29 '17 at 1:10
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It's a fact of limits that if $\lim_{x\to a} g(x)$ exists and $\lim_{x\to a} h(x)$, then $\lim_{x \to a} g(x)h(x)$ exists and is equal to the product $(\lim_{x\to a}g(x))(\lim_{x\to a}(h(x))$. That is the fact you are using for both problems. In the first case, you have $g(x)=\frac{f(x)}{x^2}$ and $h(x)=x^2$; in the second problem you have $g(x)=\frac{f(x)}{x^2}$ and $h(x)=x$.

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