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6 votes

Limit of $\lim\limits_{x \to -3} \frac {4^\frac{x+3}{5}-1}{x+3}$, without L'Hopital's rule.

HINT Are you acquainted to the derivative definition? \begin{align*} \lim_{x\to-3}\frac{4^{\frac{x+3}{5}} - 1}{x + 3} = \lim_{x\to-3}\frac{4^{\frac{x+3}{5}} - 4^{\frac{-3 + 3}{5}}}{x - (-3)} \end{...
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3 votes

Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$

Hint Take the natural logarithm and use that $z-\frac{z^2}2\le \ln(1+z) \le z$. Additionally we can switch to radial coordinates $x=r\cos\theta,\, y=r\sin\theta$ and let $r\to 0$.
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1 vote

Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$

An answer without polar coordinates, using bounds as OP was interested in. The Bernoulli inequality yields $(1+x^2y^2)^{(x^2+y^2)^{-1}}\ge1+\frac{x^2y^2}{x^2+y^2}$ so we can straightaway bound: $$\...
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1 vote

Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$

An alternative using polar coordinates: Taking $x=r\cos \theta\;,y=r\sin\theta$ $$\lim_{r\to 0}(1+r^4\sin^2\theta\cos^2\theta)^{-1/r^2}=e^{\lim_{r\to 0}\frac{-1}{r^2}(r^4\sin^2\theta\cos^2\theta)}=e^{\...
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1 vote

Calculate $\lim_{x \to 0} \frac{\cos(x)}{\sin(x)}$

I would give you full score without any hesitation. Some professors would ask for an $\epsilon$-$\delta$ proof but in my opinion this would be counter productive. Here are some reasons why I like your ...
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  • 1,439
1 vote

Limit of $\lim\limits_{x \to -3} \frac {4^\frac{x+3}{5}-1}{x+3}$, without L'Hopital's rule.

If you prefer a proof without using derivatives: Observe that our limit is also equal to $$\lim\limits_{x\to-3}\frac {4^{(x+3)/5}-1}{x+3}=\frac 15\lim\limits_{x\to0}\frac {4^x-1}x$$ Where the ...
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  • 4,547
1 vote
Accepted

Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$

You can also observe that $$ \begin{array}{l} \left( {1 + x^2 y^2 } \right)^{ - \frac{1}{{x^2 + y^2 }}} = e^{ - \frac{1}{{x^2 + y^2 }}\log \left( {1 + x^2 y^2 } \right)} = \\ \\ = e^{ - \...
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