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 limx0[a2+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: x0+x\to 0^{+}. Here 1x+\frac1x\to+\infty, so e1/x+e^{1/x}\to+\infty and 2+e1/x1+e1/x1\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/xe(1+x)^{1/x}\to e. Hence L+=a+eL_{+}=a+e.

  3. Left-hand side: x0x\to 0^{-}. Now 1x\frac1x\to-\infty, so e1/x0e^{1/x}\to 0 and 2+e1/x1+e1/x21=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 (1x)1/x(1-x)^{1/x}. Putting t=x0+t=-x\to 0^{+} gives (1+t)1/te1(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=e1e.a+e=2a+\frac1e\ \Longrightarrow\ a=e-\frac1e.

  5. Check the common value. With a=e1e2.35040a=e-\tfrac1e\approx 2.35040, the right limit is a+e5.06868a+e\approx 5.06868 and the left limit is 2a+1e4.70081+0.367885.068682a+\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=e1ea=e-\frac{1}{e}

Need to solve a different problem like this? Open the solver →