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You wanted an \varepsilon\delta proof but the claim is wrong So I will provide an \varepsilon\delta that the claim is wrong More specifically I will show that \lim _{x \to 0} \frac {sin \frac {1} {x }} {sin \frac {1} {x }}=1.
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For the calculation result of a limit such as the following `lim _(x >0) sin (x )/x ` enter limit (`sin (x )/x x `) Calculating the limit at plus infinity of a function It is possible to calculate the limit at + infini of a function If the limit exists and that the calculator is.
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Find $\lim _{x \to 0^+}\sin(x )\ln(x )$ By using l’Hôpital rule because we will get $0\times\infty$ when we substitute I rewrote it as $$\lim _{x \to0^+}\dfrac{\sin(x )}{\dfrac1{\ln(x )}}$$ to get the form $\dfrac 00$ Then I differentiated the numerator and denominator and I got $$\dfrac{\cos x }{\dfrac{1}{x (\ln.
