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Calculus · real student question

The limit as x approaches 0 of a times (2 + e^(1/x))/(1 + e^(1/x)) plus (1 + |x|)^(1/x) exists. Find the value of a.

Question

Given that lim⁡x→0[a⋅2+e1/x1+e1/x+(1+∣x∣)1/x]\lim_{x\to 0}\left[a\cdot\frac{2+e^{1/x}}{1+e^{1/x}}+\left(1+|x|\right)^{1/x}\right] exists, find aa.

Step-by-step solution

  1. Recognise why the limit is two-sided trouble. Both e1/xe^{1/x} and ∣x∣|x| behave differently on the two sides of 00, so the limit exists precisely when the right-hand and left-hand limits are equal. Compute each side separately and then force them to match.

  2. Right-hand side: x→0+x\to 0^{+}. Here 1x→+∞\frac1x\to+\infty, so e1/x→+∞e^{1/x}\to+\infty and 2+e1/x1+e1/x→1\frac{2+e^{1/x}}{1+e^{1/x}}\to 1 (divide top and bottom by e1/xe^{1/x}). Also ∣x∣=x|x|=x, so the second term is (1+x)1/x→e(1+x)^{1/x}\to e. Hence L+=a+eL_{+}=a+e.

  3. Left-hand side: x→0−x\to 0^{-}. Now 1x→−∞\frac1x\to-\infty, so e1/x→0e^{1/x}\to 0 and 2+e1/x1+e1/x→21=2.\frac{2+e^{1/x}}{1+e^{1/x}}\to\frac{2}{1}=2. Also ∣x∣=−x|x|=-x, so the second term is (1−x)1/x(1-x)^{1/x}. Putting t=−x→0+t=-x\to 0^{+} gives (1+t)−1/t→e−1(1+t)^{-1/t}\to e^{-1}. Hence L−=2a+1eL_{-}=2a+\dfrac1e.

  4. Set the two one-sided limits equal. a+e=2a+1e ⟹ a=e−1e.a+e=2a+\frac1e\ \Longrightarrow\ a=e-\frac1e.

  5. Check the common value. With a=e−1e≈2.35040a=e-\tfrac1e\approx 2.35040, the right limit is a+e≈5.06868a+e\approx 5.06868 and the left limit is 2a+1e≈4.70081+0.36788≈5.068682a+\tfrac1e\approx 4.70081+0.36788\approx 5.06868. The two agree, so the limit does exist for this aa and for no other.

Answer

a=e−1ea=e-\frac{1}{e}

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