I was given to solve the following limit with and without l'hospital's rule, I've managed to solve it without by putting the substitution 1/x=t... I don't know how to solve it by l'hospital. Please help. The limit is :

$$\lim_{x \to \infty} \frac{2x^3+1}{x-2}\cdot \sin\left(\frac{1}{x^2}\right)$$

  • $\begingroup$ please use $\LaTeX$ to help to make this readable? $\endgroup$ – Dr. Sonnhard Graubner Sep 24 '15 at 17:23
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    $\begingroup$ When dealing with a $0 \cdot \infty$ indeterminate form, take the reciprocal of one of the factors to convert it to $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Then apply L'Hôpital's Rule as usual. $\endgroup$ – Matthew Leingang Sep 24 '15 at 17:25

Let $\displaystyle x = \frac{1}{y}\;,$ Then when $\displaystyle x\rightarrow \infty\;,$ Then $y\rightarrow 0$

So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{(2+y^3)\sin y^2}{y^2(1-2y)} = \lim_{y\rightarrow 0}\frac{(y^3+2)\sin y^2}{y^2-2y^3}\left(\frac{0}{0}\right)$$ form

Now Using $\bf{L\; Hopital \; Rule}$

So $$\displaystyle \lim_{y\rightarrow 0}\frac{(y^3+2)\cos y^2\cdot 2y+\sin y^2\cdot (3y^2)}{2y-6y^2} = \lim_{y\rightarrow 0}\frac{(y^3+2)\cdot 2\cos y^2+\sin y^2\cdot 3y}{2-6y}$$

So we get limit $$\displaystyle = \frac{4+0}{2-0} = 2$$

  • $\begingroup$ nice solution $2$ is the right answer $\endgroup$ – Dr. Sonnhard Graubner Sep 24 '15 at 17:37

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